Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing3 minutes read (About 436 words)

Bestcoder

Solved Problems
Numbers
Game
Subtrees

Numbers

可能需要两个特判？

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/**
 * Oct 31, 2015 7:01:33 PM
 * PrjName: Bc61-01
 * @semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    static int maxl=1001;
    static int[] f=new int[maxl];
    static Vector<Integer> v=new Vector<Integer>();
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        while(in.nextLine()!=null){
            int n=in.nextInt();
//            if (n<1) continue;
            boolean has1=false,has2=false,has=false;;
            Arrays.fill(f, 0);
            for(int i=1;i<=n;i++){
                int k=in.nextInt();
                f[k]++;
                if (f[k]>1&&k>0) has1=true;
                if (f[k]>2) has2=true;
            }
            v.clear();
            for(int i=0;i<maxl;i++)
                if (f[i]>0)
                    v.add(i);
//            for(Integer e:v.toArray(new Integer[0]))
//                out.print(e+" ");
            if (v.get(0)==0&&has1||has2){
                out.println("YES");continue;
            }
            Integer[] a=v.toArray(new Integer[0]);
            for(int i=0;i<a.length-1;i++)
                if (!has)
                    for(int j=a.length-1;j>i;j--){
                        int k=Arrays.binarySearch(a, a[j]-a[i]);
                        if (k>i&&k<j){
                            has=true;break;
                        }
                            
                    }
            out.println(has?"YES":"NO");
        }
        out.flush();
        out.close();
    }
}

Game

各种分类讨论，但就是要写好最边缘的特判。。。

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/**
 * Oct 31, 2015 7:28:53 PM
 * PrjName: Bc61-02
 * @semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        while(in.nextLine()!=null){
            int n=in.nextInt();
            int s=in.nextInt();
            int t=in.nextInt();
            if (s==t){
                out.println(n>1?-1:0);continue;
            }
            if (t==1){
                if (s==n)
                    out.println(0);
                else if (s==2)
                    out.println(1);
                else 
                    out.println(2);
            }
            else if (t==n){
                if (s==1)
                    out.println(0);
                else if (s==n-1)
                    out.println(1);
                else
                    out.println(2);
            }
            else{
                if (s==t+1||s==t-1||s==1||s==n)
                    out.println(1);
                else 
                    out.println(2);
            }
        }
        out.flush();
        out.close();
    }

}

Subtrees

有些查找行为显然拖慢时间以致TLE。。。
有一个重要优化是从小到大枚举子树大小而不是相反方向；

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/**
 * Nov 7, 2015 10:31:31 PM
 * PrjName: hdu5524-3
 * @semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    static int maxn=64;
    static long[] p2;
    static HashSet<Long> st=new HashSet<Long>();
    static PrintWriter out=new PrintWriter(System.out);
    static void solve(long n){
        if (n==0L) return;
        st.add(n);
        if (n==1L) return;
        long f;
        for(f=1L;;f=f<<1|1L)
            if ((f>>1)<=n-1L-f&&(f<<1)>=n-1L-f)
                break;
        solve(f);
        if ((f<<1|1)!=n)
            solve(n-1-f);
    }
    static void init(){
        p2=new long[maxn];
        p2[0]=1L;
        for(int i=1;i<maxn;i++){
            p2[i]=p2[i-1]<<1;
            p2[i-1]--;
        }
    }
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        
        init();
        while(in.nextLine()!=null){
            long n=in.nextLong();
            st.clear();solve(n);
            out.println(st.size());
        }
        out.flush();
        out.close();
    }

}
Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing3 minutes read (About 410 words)

BestCoder

Solved Problems
SDOI
Reorder the Books

SDOI

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/**
 * 2015年10月10日 下午6:54:59
 * PrjName:Bc59-01
 * @ Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    static Person[] ps;
    static ArrayList<Person> li=new ArrayList<Person>();
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int T=in.nextInt();
        while(T-->0){
            int n=in.nextInt();
            int m=in.nextInt();
            float x1=0,x2=0;
            int female=0,h=0;
            ps=new Person[n];
            for(int i=0;i<n;i++){
                ps[i]=new Person(in.next(), in.next(), in.nextInt(), in.nextInt());
                female+=ps[i].sex;
                if (ps[i].sex==1&&h==0) h=i;
                x1=Math.max(x1, ps[i].s1);
                x2=Math.max(x2, ps[i].s2);
            }
            li.clear();
            for(int i=0;i<n;i++){
                ps[i].s1*=300;ps[i].s1/=x1;
                ps[i].s2*=300;ps[i].s2/=x2;
                ps[i].s=ps[i].s1*(float)0.3+ps[i].s2*(float)0.7;
                if (female>0){
                    if (ps[i].s>ps[h].s&&ps[i].sex==1) h=i;
                }
                else
                    if (ps[i].s>ps[h].s) h=i;
                //li.add(ps[i]);
            }
            out.println("The member list of Shandong team is as follows:");
            Person p0=new Person();
            if (female>0){
//                out.println(li.get(h).name);
                p0=new Person(ps[h]);
                ps[h].s=-1;
                m--;
            }
            Arrays.sort(ps, new Person.comp());
            li.clear();
            
            for(int i=0;i<m;i++)
                li.add(ps[i]);
            if (female>0){
                li.add(p0);m++;
            }
            ps=li.toArray(new Person[0]);
            Arrays.sort(ps, new Person.comp());
            for(int i=0;i<m;i++)
                out.println(ps[i].name);

        }
        out.flush();
        out.close();
    }
}
class Person{
    String name;
    int sex;
    float s1,s2,s;
    Person(){}
    Person(Person p){
        name=new String(p.name);
        sex=p.sex;
        s1=p.s1;
        s2=p.s2;
        s=p.s;
    }
    Person(String nm,String sx,int _s1,int _s2){
        name=new String(nm);
        if (sx.charAt(0)=='m') sex=0;
        else sex=1;
        s1=_s1;s2=_s2;
    }
    void debug(PrintWriter out){
        out.println(name+"\t"+sex+"\t"+s);
    }
    static class comp implements Comparator<Person> {
        public int compare(Person p1,Person p2){
            return -Float.compare(p1.s, p2.s);
        }
    }
}

Reorder the Books

思索再三，竟然连充分条件、必要条件都没找准，

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/** Oct 13, 2015 8:55:41 PM
 * PrjName:hdu5500
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    static int[] a;
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int T=in.nextInt();
        while(T-->0){
            int n=in.nextInt();
            a=new int[n+1];
            for(int i=1;i<=n;i++)
                a[in.nextInt()]=i;
            int ans=0;
//            if (a[n]!=n&&a[n]!=1) ans++;
            int k=a[n];
            for(int i=n;i>1;i--)
                if (a[i-1]<k){
                    k=a[i-1];n--;
                }
                else 
                    break;
//            out.println(n);    
            for(int i=n-1;i>1;i--){
                if (a[i]==1)
                    continue;
                boolean need=false;
                for(int j=1;j<i;j++)
                    if (a[j]>a[i]) need|=true;
                for(int j=i+1;j<=n;j++)
                    if (a[j]<a[i]) need|=true;
                if (need){
                    ans++;
                    for(int j=1;j<=n;j++)
                        if (a[j]<a[i])
                            a[j]++;
                    a[i]=1;
                }
            }
            if (a[1]!=1) ans++;
            out.println(ans);
        }
        out.flush();
        out.close();
    }
}

观察、归纳后可以优化到最简，

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public class Main {
    static int[] a;
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int T=in.nextInt();
        while(T-->0){
            int n=in.nextInt();
            a=new int[n+1];
            for(int i=1;i<=n;i++)
                a[in.nextInt()]=i;
            int ans=0;
            int k=a[n],h=0;
            for(int i=n;i>1;i--){
                if (a[i-1]>k){
                    ans++;
                    a[i-1]=h--;
                }
                k=a[i-1];
            }
            if (a[1]!=h+1) ans++;
            out.println(ans);
        }
        out.flush();
        out.close();
    }

}
Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing2 minutes read (About 249 words)

BestCoder

1001 Clarke and minecraft

玩过MC无压力 

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/** Sep 19, 2015 7:12:36 PM
 * PrjName:Bc56-01
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {

    /**
     * @param args
     */
    final static int maxn=110;
    static int cnt;
    static int[] w;
    static HashMap<Integer, Integer> mp=new HashMap<Integer,Integer>();
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int T=in.nextInt();
        //h=new int[maxn];
        w=new int[maxn];
        while(T-->0){
            int n=in.nextInt();
            cnt=0;
            mp.clear();
            Arrays.fill(w, 0);
            for(int i=1;i<=n;i++){
                int k=in.nextInt();
                if (mp.containsKey(k))
                    w[mp.get(k)]+=in.nextInt();
                else{
                    w[cnt]+=in.nextInt();
                    mp.put(k, cnt++);
                }
            }
            int sum=0;
            for(int i=0;i<cnt;i++)
                sum+=(w[i]+63)/64;
            out.println((sum+35)/36);
        }
        out.flush();
        out.close();
    }
}

1002 Clarke and problem

第一时间想到的是容斥，而后才逐渐醒悟过来是DP。。。
另外mod的遗漏又错了一发。。。

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/** Sep 19, 2015 7:24:54 PM
 * PrjName:Bc56-02
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {

    /**
     * @param args
     */
    final static int maxn=1010;
    final static int high=1000000000;
    final static int mod=1000000007;
    //static int[] s=new int[maxn];
    static int[][] f=new int[2][maxn];
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int T=in.nextInt();
        while(T-->0){
            int n=in.nextInt();
            int p=in.nextInt();
            //Arrays.fill(s, 0, p, 0);
            Arrays.fill(f[0], 0, p, 0);
            f[0][0]=1;
            int cnt=0;
            //for(int j=0;j<p;j++)out.print(f[j]+" ");out.println();
            for(int i=1;i<=n;i++){
                int k=in.nextInt();
                while(k<0) k+=p;
                k%=p;
                Arrays.fill(f[cnt^1], 0,p,0);
                for(int j=0;j<p;j++){
                    f[cnt^1][(k+j)%p]+=f[cnt][j];
                    f[cnt^1][(k+j)%p]%=mod;
                }
                for(int j=0;j<p;j++){
                    f[cnt^1][j]+=f[cnt][j];
                    f[cnt^1][j]%=mod;
                }
                cnt^=1;
            }
            out.println(f[cnt][0]);
            
        }
        out.flush();
        out.close();
    }
}

1003 Clarke and puzzle

没有总结好Nim游戏的规律啊。。。超简单的代码

Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing4 minutes read (About 640 words)

Bestcoder

1001 A problem of sorting

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/** Sep 5, 2015 7:07:52 PM
 * PrjName:Bc54-01
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    final static int maxn=116;
    static String s;
    static Vector<String>[] vec=new Vector[maxn];
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        for(int i=0;i<maxn;i++)
            vec[i]=new Vector<String>();
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int T=Integer.parseInt(br.readLine());
        while(T-->0){
            for(int i=0;i<maxn;i++)
                vec[i].clear();
            int n=Integer.parseInt(br.readLine());
            for(int i=1;i<=n;i++){
                s=new String(br.readLine());
                int y=0,tmp=1,j;
                for(j=s.length()-1;j>0;j--)
                    if (s.charAt(j)>='0'&&s.charAt(j)<='9'){
                        y+=tmp*(s.charAt(j)-'0');
                        tmp*=10;
                    }
                    else 
                        break;
                s=s.substring(0, j);
                vec[y-1900].add(s);
            }
            for(int i=maxn-1;i>=0;i--)
                for(int j=0;j<vec[i].size();j++)
                    out.println(vec[i].get(j));
        }
        out.flush();
        out.close();
    }
}

1002 The Factor

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/** Sep 5, 2015 7:30:23 PM
 * PrjName:Bc54-02
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    final static int maxn=109000;
    static int[] a;
    static int[] pri,phi,fstp;
    static void get_prime(){
        pri=new int[maxn];
        fstp=new int[maxn];
        phi=new int[maxn];
        phi[1]=1;
        for(int i=2;i<maxn;i++){
            if (fstp[i]==0){
                pri[++pri[0]]=i;
                phi[i]=i-1;
            }
            for(int j=1;j<=pri[0]&&i*pri[j]<maxn;j++){
                int k=i*pri[j];
                fstp[k]=pri[j];
                //if (fstp[i]==pri[j]){
                if (i%pri[j]==0){
                    phi[k]=phi[i]*pri[j];
                    break;
                }
                else
                    phi[k]=phi[i]*(pri[j]-1);
            }
        }
    }
    static int get_fstp(int n){
        if (n<maxn)
            return fstp[n]>0?fstp[n]:n;
        for(int i=2;i<maxn;i++)
            if (n%i==0)
                return fstp[i]>0?fstp[i]:i;
        return 0;
    }
    static Vector<Integer> get_prime_factor(int n){
        Vector<Integer> res=new Vector<Integer>();
        while(n>1&&fstp[n]>0){
            res.add(fstp[n]);
            n/=fstp[n];
        }
        if (n>1) res.add(n);
        return res;
    }
    static Vector<Integer> v=new Vector<Integer>();
    static Integer[] f;
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        get_prime();
        int T=in.nextInt();
        while(T-->0){
            int n=in.nextInt();
            v.clear();
            a=new int[n+1];
            for(int i=1;i<=n;i++){
                a[i]=in.nextInt();
                if (a[i]==1) continue;
                int tmp=get_fstp(a[i]);
                v.add(tmp);
                if (a[i]/tmp>1) v.add(get_fstp(a[i]/tmp));
            }
            /*v.sort(new Comparator<Integer>() {
                public int compare(Integer o1,Integer o2){
                    return Integer.compare(o1, o2);
                }
            });*/
            f=v.toArray(new Integer[0]);
            Arrays.sort(f);
            if (v.size()<2)
                out.println(-1);
            else 
                out.println(f[0]*f[1]);
        }
        out.flush();
        out.close();
    }
}

⬆️当时想太多了啊，何必把找到的因子全存进vector啊，白白吃了个RE……
（测试还发现HDU尚不支持Vector的sort方法，还未支持到Java7？）

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/** Sep 8, 2015 16:02:23 PM
 * PrjName:Bc54-02
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    final static int maxn=45000;
    static int[] a;
    static int[] pri,fstp;
    static void get_prime(){
        pri=new int[maxn];
        fstp=new int[maxn];
        for(int i=2;i<maxn;i++){
            if (fstp[i]==0){
                pri[++pri[0]]=i;
            }
            for(int j=1;j<=pri[0]&&i*pri[j]<maxn;j++){
                int k=i*pri[j];
                fstp[k]=pri[j];
                if (i%pri[j]==0)
                    break;
            }
        }
    }
    static int get_fstp(int n){
        if (n<maxn)
            return fstp[n]>0?fstp[n]:n;
        for(int i=2;i<maxn;i++)
            if (n%i==0)
                return fstp[i]>0?fstp[i]:i;
        return n;
    }
    static int max1,max2,sum;
    static void init(){
        max1=max2=0x3fffffff;sum=0;
    }
    static void update(int s){
        if (s<max1){
            max2=max1;max1=s;
        }
        else 
            max2=Math.min(max2, s);
        sum++;
    }
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        get_prime();
        int T=in.nextInt();
        while(T-->0){
            init();
            int n=in.nextInt();
            a=new int[n+1];
            for(int i=1;i<=n;i++){
                a[i]=in.nextInt();
                if (a[i]==1) continue;
                int tmp=get_fstp(a[i]);
                update(tmp);
                if (a[i]/tmp>1) update(get_fstp(a[i]/tmp));
            }
            if (sum<2)
                out.println(-1);
            else 
                out.println(max1*(long)max2);
        }
        out.flush();
        out.close();
    }
}

1003 Geometric Progression

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/** Sep 5, 2015 8:05:33 PM
 * PrjName:Bc54-03
 * @author Semprathlon
 */
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
    final static int maxn=101;
    static BigInteger a[]=new BigInteger[maxn];
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int T=in.nextInt();
        while(T-->0){
            int n=in.nextInt();
            boolean zero=false,nonzero=false;
            for(int i=1;i<=n;i++){
                a[i]=new BigInteger(in.next());
                if (a[i].equals(BigInteger.ZERO))
                    zero|=true;
                else
                    nonzero|=true;
            }
            boolean ans=(zero&&nonzero)?false:true;
            if (ans)
            for(int i=2;i<n;i++)
                if (!a[i].pow(2).equals(a[i-1].multiply(a[i+1]))){
                    ans=false;
                    break;
                }
            out.println(ans?"Yes":"No");
        }
        out.flush();
        out.close();
    }
}

不明不白地送了个WA，没有考虑好数列中值为0的项。
这题不是暑假多校才做过的么？

Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing5 minutes read (About 742 words)

Bestcoder

迟到了半小时开打，不然rank还可以更好看点……

1001 Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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/** Aug 29, 2015 7:39:29 PM
 * PrjName:Bc53-01
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        StreamTokenizer in=new StreamTokenizer(new BufferedInputStream(System.in));
        PrintWriter out=new PrintWriter(System.out);
        while(in.nextToken()!=StreamTokenizer.TT_EOF){
            int n=(int)in.nval;
            in.nextToken();
            int m=(int)in.nval;
            boolean has=false;
            //out.println(n+"-"+m);
            for(int i=1;i<=m;i++){
                in.nextToken();
                int u=(int)in.nval;
                in.nextToken();
                int v=(int)in.nval;
                if (u==1&&v==n||u==n&&v==1)
                    has=true;
            }
            if (has)
                out.println(1+" "+n*(n-1)/2);
            else
                out.println(1+" "+1);
        }
        out.flush();
        out.close();
    }
}

如果点1与点n直接相连，那么就是距离最短了。

1002 Rikka with Tree

要同时满足不同与相似，就是要求结点数相等，各结点到树根1的距离分别相等（即深度分别相同），且存在父节点不同的结点。
可以说就是判断同一深度上的结点互换后是否能得到新的结构。
若有某一深度上的结点数多于一个，那么大概可以通过同层互换获得新的结构。

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/** Aug 29, 2015 7:59:14 PM
 * PrjName:Bc53-02
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {

    /**
     * @param args
     */
    final static int maxn=2010;
    static Graph G=new Graph(maxn);
    static int[] dep;
    static HashSet<Integer> st=new HashSet<Integer>();
    static void bfs(int st){
        Queue<Integer> que=new LinkedList<Integer>();
        que.add(st);
        while(!que.isEmpty()){
            int u=que.poll();
            for(int i=G.h[u];i>-1;i=G.edge[i].next){
                int v=G.edge[i].to;
                if (dep[v]>0) continue;
                dep[v]=dep[u]+1;
                que.add(v);
            }
        }
    }
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        StreamTokenizer in=new StreamTokenizer(new BufferedInputStream(System.in));
        PrintWriter out=new PrintWriter(System.out);
        while(in.nextToken()!=StreamTokenizer.TT_EOF){
            int n=(int)in.nval;
            G.clear();
            dep=new int[n+1];
            dep[1]=1;
            for(int i=1;i<n;i++){
                in.nextToken();
                int u=(int)in.nval;
                in.nextToken();
                int v=(int)in.nval;
                G.add(u, v);
                G.add(v, u);
            }
            bfs(1);
            boolean ans=true;
            st.clear();
            for(int i=1;i<=n;i++)
                if (st.contains(dep[i])){
                    ans=false;
                    break;
                }
                else
                    st.add(dep[i]);
            out.println(ans?"YES":"NO");
        }
        out.flush();
        out.close();
    }

}
class Edge{
    int to,next;
    Edge(int _u,int _v){
        to=_u;next=_v;
    }
}
class Graph{
    int[] h;
    int sz;
    Edge[] edge;
    Graph(int size){
        sz=size;
        h=new int[sz+1];
        edge=new Edge[sz+1];
        Arrays.fill(h, -1);
        h[0]=0;
    }
    void clear(){
        h=new int[sz+1];
        edge=new Edge[sz+1];
        Arrays.fill(h, -1);
        h[0]=0;
    }
    void add(int u,int v){
        edge[h[0]]=new Edge(v,h[u]);
        h[u]=h[0]++;
    }
}

然而错了。
适当地列举一些情况，发现若树的深度仅有2，那么它始终是特殊的。
可是加上这一判断仍不够。
进一步的举例发现，若树上非最底层有某一层结点多于一个，就能变换形成新结构

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/** Aug 29, 2015 7:59:14 PM
 * PrjName:Bc53-02
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {

    /**
     * @param args
     */
    final static int maxn=2010;
    static Graph G=new Graph(maxn);
    static int[] dep;
    static HashSet<Integer> st=new HashSet<Integer>();
    static void bfs(int st){
        Queue<Integer> que=new LinkedList<Integer>();
        que.add(st);
        while(!que.isEmpty()){
            int u=que.poll();
            for(int i=G.h[u];i>-1;i=G.edge[i].next){
                int v=G.edge[i].to;
                if (dep[v]>0) continue;
                dep[v]=dep[u]+1;
                que.add(v);
            }
        }
    }
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        StreamTokenizer in=new StreamTokenizer(new BufferedInputStream(System.in));
        PrintWriter out=new PrintWriter(System.out);
        while(in.nextToken()!=StreamTokenizer.TT_EOF){
            int n=(int)in.nval;
            G.clear();
            dep=new int[n+1];
            dep[1]=1;
            for(int i=1;i<n;i++){
                in.nextToken();
                int u=(int)in.nval;
                in.nextToken();
                int v=(int)in.nval;
                G.add(u, v);
                G.add(v, u);
            }
            bfs(1);
            boolean ans=true;
            st.clear();
            int maxd=0;
            for(int i=1;i<=n;i++)
                maxd=Math.max(dep[i], maxd);
            for(int i=1;i<=n;i++)
                if (st.contains(dep[i])){
                    if (dep[i]<maxd){
                        ans=false;
                        break;
                    }
                }
                else
                    st.add(dep[i]);
            if (maxd<3)
                ans=true;
            out.println(ans?"YES":"NO");
        }
        out.flush();
        out.close();
    }

}
class Edge{
    int to,next;
    Edge(int _u,int _v){
        to=_u;next=_v;
    }
}
class Graph{
    int[] h;
    int sz;
    Edge[] edge;
    Graph(int size){
        sz=size;
        h=new int[sz+1];
        edge=new Edge[sz+1];
        Arrays.fill(h, -1);
        h[0]=0;
    }
    void clear(){
        h=new int[sz+1];
        edge=new Edge[sz+1];
        Arrays.fill(h, -1);
        h[0]=0;
    }
    void add(int u,int v){
        edge[h[0]]=new Edge(v,h[u]);
        h[u]=h[0]++;
    }
}

这与题解中所说的“一棵树是独特的当且仅当任意处于每一个深度的点数是1 1 1 1 … 1 1 x”相符（我个人将这种形态称作“呈扫帚状”）

Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing2 minutes read (About 338 words)

BestCoder

1001 Victor and Machine

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

  • Problem Description
    Victor has a machine. When the machine starts up, it will pop out a ball immediately. After that, the machine will pop out a ball every w seconds. However, the machine has some flaws, every time after x seconds of process the machine has to turn off for y seconds for maintenance work. At the second the machine will be shut down, it may pop out a ball. And while it’s off, the machine will pop out no ball before the machine restart.
    Now, at the 0 second, the machine opens for the first time. Victor wants to know when the n-th ball will be popped out. Could you tell him?

  • Input
    The input contains several test cases, at most 100 cases.
    Each line has four integers x, y, w and n. Their meanings are shown above。
    $ 1\leq x,y,w,n\leq 100 $ .

  • Output
    For each test case, you should output a line contains a number indicates the time when the n-th ball will be popped out.

  • Sample Input
    2 3 3 3
    98 76 54 32
    10 9 8 100

  • Sample Output
    10
    2664
    939

n-1的问题，自己刚开始出了点差错：

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/** Aug 22, 2015 7:03:37 PM
 * PrjName:Bc52-01
 * @author Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        StreamTokenizer in=new StreamTokenizer(new BufferedInputStream(System.in));
        PrintWriter out=new PrintWriter(System.out);
        while(in.nextToken()!=StreamTokenizer.TT_EOF){
            int x=(int)in.nval;
            in.nextToken();
            int y=(int)in.nval;
            in.nextToken();
            int w=(int)in.nval;
            in.nextToken();
            int n=(int)in.nval;
            if (n==1){
                out.println(0);continue;
            }
            if (w>x){
                out.println((n-1)*(x+y));continue;
            }
            else if (w==x){
                out.println((n-1)/2*(x+y)+(n-1)%2*x);continue;
            }
            else{
                int tmp=(x+w)/w;
                out.println((n-1)/tmp*(x+y)+(n-1)%tmp*w);continue;
            }
        }
        out.flush();
        out.close();
    }

}
Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing3 minutes read (About 466 words)

BestCoder

1001 Distribution money

WA过一发，因为忽视了金额的分布范围。

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/**
 * 2015年8月8日 下午7:02:20
 * PrjName:Bc50-01
 * @ Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    static int[] a;
    static int maxn=10001;
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        StreamTokenizer in=new StreamTokenizer(new BufferedInputStream(System.in));
        PrintWriter out=new PrintWriter(System.out);
        a=new int[maxn];
        while(in.nextToken()!=StreamTokenizer.TT_EOF){
            int n=(int)in.nval;
            Arrays.fill(a, 0);
            for(int i=1;i<=n;i++){
                in.nextToken();
                int k=(int)in.nval;
                a[k]++;
            }
            int ans=-1;
            for(int i=0;i<maxn;i++)
                if ((a[i]<<1)>n)
                    ans=i;
            //if (n>1)
                out.println(ans);
            //else 
                //out.println(-1);
        }
        out.flush();
        out.close();
    }
}

1002 Run

明显搞复杂了，做了一回出题人眼中的“火星人”T_T
平面上的整点就是无法构成正三角形、正五边形、正六边形，没这点见识就只有瞎弄。。。
简化到这个地步，正方形的判断就应该仔细点了吧？四条边及两对角线的长度比较都写上， 

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/**
 * Nov 19, 2015 7:32:18 PM
 * PrjName: hdu5365
 * @semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    static Point[] pt;
    static boolean check(Point p1,Point p2,Point p3,Point p4){
        double[] v=new double[6];
        v[0]=Point.dist(p1, p2);
        v[1]=Point.dist(p2, p3);
        v[2]=Point.dist(p3, p4);
        v[3]=Point.dist(p1, p4);
        v[4]=Point.dist(p1, p3);
        v[5]=Point.dist(p2, p4);
        Arrays.sort(v);
        return v[0]==v[1]&&v[1]==v[2]&&v[2]==v[3]&&v[4]==v[5];
    }
    public static void main(String[] args) throws IOException{
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        while(in.nextLine()!=null){
            int n=in.nextInt();
            pt=new Point[n];
            for(int i=0;i<n;i++)
                pt[i]=new Point(in.nextInt(), in.nextInt());
            int ans=0;
            for(int i=0;i<n;i++)
                for(int j=i+1;j<n;j++)
                    for(int k=j+1;k<n;k++)
                        for(int l=k+1;l<n;l++){
                            Point p1=pt[i],p2=pt[j],p3=pt[k],p4=pt[l];
                            if (check(p1, p2, p3, p4))
                                ans++;
                        }
            out.println(ans);
        }
        out.flush();
        out.close();
    }
}

1003 The mook jong

多么有教育意义的猜公式，猜不出就别偏执了。。。
f[i]=s[i-3]+1,s[i]=s[i-1]+f[i],…

Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing3 minutes read (About 467 words)

Bestcoder

1001 Untitled

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) 

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/**
 * 2015年8月1日 下午7:15:14
 * PrjName:Bc49-01
 * @ Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    static ArrayList<Integer> v=new ArrayList<Integer>();
    static int[] a,f;
    static int lowbit(int x){
        return x&(-x);
    }
    static int get(int x){
        int res=0;
        while(x>0){
            res++;
            x-=lowbit(x);
        }
        return res;
    }
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int T=in.nextInt();
        while(T-->0){
            v.clear();
            int n=in.nextInt();
            int m=in.nextInt();
            for(int i=1;i<=n;i++)
                v.add(in.nextInt());
            Collections.sort(v);
            f=new int[1<<n];
            Arrays.fill(f, -1);
            f[0]=m;
            //out.println(n);
            int ans=Integer.MAX_VALUE;
            for(int i=0;i<(1<<n);i++)
                if (f[i]>0)
                    for(int j=0;j<n;j++)
                        if (((1<<j)&i)==0){
                            int k=i|(1<<j);
                            f[k]=f[i]%v.get(j);
                            if (f[k]==0)
                                ans=Math.min(ans, get(k));
                        }
            if (ans==Integer.MAX_VALUE) out.println(-1);
            else out.println(ans);
        }
        out.flush();
        out.close();
    }
}

1002 Three Palindromes

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

  • Problem Description
    Can we divided a given string S into three nonempty palindromes?

  • Input
    First line contains a single integer $ T \leq 20 $ which denotes the number of test cases.
    For each test case , there is an single line contains a string S which only consist of lowercase English letters. $ 1\leq |s| \leq 20000 $

  • Output
    For each case, output the “Yes” or “No” in a single line.

  • Sample Input
    2
    abc
    abaadada

  • Sample Output
    Yes
    No

赛时的Hash做法：

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/**
 * 2015年8月1日 下午7:47:15
 * PrjName:Bc49-02
 * @ Semprathlon
 */
import java.io.*;
import java.util.*;
public class Main {
    static long pri=29L;
    static long lh(String s,long res,int p){
        return res*pri+(long)(s.charAt(p)-'a');
    }
    static long rh(String s,long res,int p,long m){
        return res+(long)(s.charAt(p)-'a')*m;
    }
    static BitSet v=new BitSet(20010);
    public static void main(String[] args) throws IOException{
        // TODO Auto-generated method stub
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int T=in.nextInt();
        while(T-->0){
            String s=new String(in.next());
            int len=s.length();
            //s=s+"#"+(new StringBuffer(s).reverse().toString());
            //out.println(s);
            boolean ans=false;
            long l1=0L,r1=0L,tmp=1L;
            for(int i=0;i<len;i++){
                if (ans) break;
                l1=lh(s, l1, i);
                r1=rh(s, r1, i, tmp);
                //out.println(i+" "+l1+" "+r1);
                tmp*=pri;
                //if ((i&1)>0) continue;
                if (l1==r1){
                    //out.println("f"+i);
                    long l2=0L,r2=0L,tmp2=1L;
                    v.clear();
                    for(int j=i+1;j<len;j++){
                        l2=lh(s,l2,j);
                        r2=rh(s,r2,j,tmp2);
                        tmp2*=pri;
                        //if (((j-i+1)&1)>0) continue;
                        if (l2==r2) v.set(j);
                    }
                    
                    l2=0L;r2=0L;tmp2=1L;
                    for(int j=len-1;j>i;j--){
                        l2=lh(s,l2,j);
                        r2=rh(s,r2,j,tmp2);
                        tmp2*=pri;
                        //if (((len-j+1)&1)>0) continue;
                        if (l2==r2&&v.get(j-1)){
                            ans=true;break;
                        }
                    }
                }
            }
            out.println(ans?"Yes":"No");
        }
        out.flush();
        out.close();
    }
}

首次尝试去hack别人不成，还被人黑了，立马TLE

题解介绍的多种方法中，个人认为二分+hash代码量较少，如何实现？

Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing2 minutes read (About 239 words)

BestCoder 1st Anniversary T_T

1001 Souvenir

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) 

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/**
 * 2015年7月25日 下午7:07:41
 * PrjName:0725-01
 * @ Semprathlon
 */
public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        InputReader in=new InputReader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int T=in.nextInt();
        while(T-->0){
            int n=in.nextInt();
            int m=in.nextInt();
            int p=in.nextInt();
            int q=in.nextInt();
            int ans=0,has=0;
            if (m*p>q)
                ans+=n/m*q;
            else ans+=n/m*m*p;
            ans+=n%m*p;
            if ((n+m)/m*q<ans) ans=(n+m)/m*q;
            out.println(ans);
        }
        out.flush();
        out.close();
    }

}

1002 Hidden String

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
用Trie做查询时，往死胡同里钻而且根本停不下来，啊啊啊啊…… 

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#include <cstdio>
#include <cstring>
char s[200], con[] = "anniversary";

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s", s);
        int len = strlen(s);
        bool flag = false;
        for(int i = 0; i <= 8; i++)
        {
            for(int j = i + 1; j <= 9; j++)
            {
                int k = 0;
                while(k < len && strncmp(con, s + k, i + 1) != 0)
                    k ++;
                if(k == len) 
                    continue;
                k += i + 1;
                while(k < len && strncmp(con + i + 1, s + k, j - i) != 0)
                    k ++;
                if(k == len) 
                    continue;
                k += j - i;
                while(k < len && strncmp(con + j + 1, s + k, 10 - j) != 0)
                    k ++;
                if(k != len)
                {
                    flag = true;
                    break;
                }
            }
        }
        if(flag) 
            puts("YES");
        else 
            puts("NO");
    }
}

这里的枚举倒是简洁得出奇。

1003 Sequence

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
下意识地做了贪心，果然被hack

Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing3 minutes read (About 416 words)

Bestcoder

Protest一时爽，final test全爆零 QwQ

1001 wyh2000 and a string problem

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out.println(str.matches(".*w.*y.*h.*")||str.matches(".*v{2,}.*y.*h.*")?"Yes":"No");

偷懒着用正则表达式，华丽地TLE……
后来就醉了一样的胡乱的写了个匹配也不太行……

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import java.io.*;
import java.util.*;
import java.math.*;
import java.util.regex.*;
import com.sun.org.apache.xalan.internal.xsltc.compiler.Pattern;
public class Main {
    static boolean check(String str){
        int res=0;
        int i,len=str.length();
        for(i=0;i<len-1;i++)
            if (str.charAt(i)=='w'||str.charAt(i)=='v'&&str.charAt(i+1)=='v'){
                res|=1;break;
            }
        for(;i<len;i++)
            if (str.charAt(i)=='y'){
                res|=2;break;
            }
        for(;i<len;i++)
            if (str.charAt(i)=='h'){
                res|=4;break;
            }
        return res==7;
    }
    public static void main(String[] args)  throws IOException{
        // TODO Auto-generated method stub
         InputReader in = new InputReader(System.in)  ;
         PrintWriter out = new PrintWriter(System.out) ;
         int T=in.nextInt();
         while(T-->0){
             String str=in.next();
             out.println(check(str)?"Yes":"No");
             out.flush();
         }
         out.close();
    }
}

1002 wyh2000 and pupil

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#include<cctype>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
const int maxn=200010;
const int inf=0x7fffffff;
const double eps=1e-3;

short col[maxn];
int s[2];

struct Edge
{
    int to,next;
    Edge(){}
    Edge(int v,int w):to(v),next(w){}
} edge[maxn];
int head[maxn];

void addedge(int u,int v)
{
    edge[head[0]]=Edge(v,head[u]);
    head[u]=head[0]++;
}

void init()
{
    fill(head,head+maxn,-1);
    head[0]=1;
}

int bfs(int u)
{
    queue<int> q;
    q.push(u);
    col[u]=0;
    while(!q.empty())
    {
        int p=q.front();
        s[col[p]]++;
        q.pop();
        for(int i=head[p];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if (col[v]>=0)
            {
                if (col[v]==col[p]) return -1;
            }
            else
            {
                col[v]=1^col[p];
                q.push(v);
            }

        }
    }
    return 0;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }

        int sum=0;
        fill(col+1,col+n+1,-1);
        for(int i=1;i<=n;i++)
            if (col[i]<0)
        {
            s[0]=s[1]=0;
            if (bfs(i)<0)
            {
                sum=-1;break;
            }
            else sum+=max(s[0],s[1]);
        }
        if (n==sum) sum--;
        if (sum<1||n-sum<1)
            puts("Poor wyh");
        else
            printf("%d %d\n",sum,n-sum);
    }
    return 0;
}

事实证明这并不是二分图匹配；并没有什么典型的算法让两个集合中的一个最大……
用DFS或BFS都可实现填色。

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