Posted Updated  Semprathlon / Simfae Dean Gita few seconds read (About 27 words)

CharacterGraph - Intro

semprathlon/CharacterGraph.

 

This Android Project is intended for an Android application to be developed, which enables mutual conversion between a picture and a matrix of characters.

Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing5 minutes read (About 714 words)

POJ1584-A Round Peg in a Ground Hole - ζёСяêτ - 小優YoU - 博客频道 - CSDN NET

POJ1584-A Round Peg in a Ground Hole - ζёСяêτ - 小優YoU - 博客频道 - CSDN.NET.

转载请注明出处：優YoU  http://user.qzone.qq.com/289065406/blog/1309142308

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//Memory Time
//268K    0MS 

#include<iostream>
#include<cmath>
using namespace std;

const double eps=1e-6;
const double pi=3.141592654;

typedef class NODE
{
	public:
		double x,y;
}pos;

int n;
double PegR;  //钉子半径
pos Peg;  //钉子坐标

int precision(double x);  //精度讨论
double det(double x1,double y1,double x2,double y2);  //叉积
double dotdet(double x1,double y1,double x2,double y2);  //点积
double cross(pos A,pos B,pos C,pos D);
double distant(pos A,pos B);  //计算距离
double angle(pos A,pos B,pos P);  //计算向量PA与PB夹角

bool IsConvexBag(pos* Vectex);  //判断输入的点集是否为凸包(本题保证了输入的点集为按某一时针方向有序)
bool IsIn(pos* Vectex); //判断圆心是否在多边形内部
bool IsFit(pos* Vectex);  //判断圆的半径是否<=其圆心到多边形所有边的最小距离

int main(void)
{
	while(cin>>n && n>=3)
	{
		cin>>PegR>>Peg.x>>Peg.y;
		pos* Vectex=new pos[n+2];  //多边形顶点坐标

		for(int i=1;i<=n;i++)
			cin>>Vectex[i].x>>Vectex[i].y;

		Vectex[0].x=Vectex[n].x;  //封闭多边形
		Vectex[0].y=Vectex[n].y;
		Vectex[n+1].x=Vectex[1].x;
		Vectex[n+1].y=Vectex[1].y;

		if(!IsConvexBag(Vectex))
			cout<<"HOLE IS ILL-FORMED"<<endl;
		else
		{
			bool flag1=IsIn(Vectex);
			bool flag2=IsFit(Vectex);

			if(flag1 && flag2)
				cout<<"PEG WILL FIT"<<endl;
			else
				cout<<"PEG WILL NOT FIT"<<endl;
		}

		delete Vectex;
	}
	return 0;
}

/*精度讨论*/
int precision(double x)
{
	if(fabs(x)<=eps)
		return 0;
	return x>0?1:-1;
}

/*计算点积*/
double dotdet(double x1,double y1,double x2,double y2)
{
	return x1*x2+y1*y2;
}

/*计算叉积*/
double det(double x1,double y1,double x2,double y2)
{
	return x1*y2-x2*y1;
}
double cross(pos A,pos B,pos C,pos D)
{
	return det(B.x-A.x , B.y-A.y , D.x-C.x , D.y-C.y);
}

/*计算距离*/
double distant(pos A,pos B)
{
	return sqrt((B.x-A.x)*(B.x-A.x)+(B.y-A.y)*(B.y-A.y));
}

/*计算角度*/
double angle(pos A,pos B,pos P)
{
	return acos(dotdet(A.x-P.x,A.y-P.y,B.x-P.x,B.y-P.y)/(distant(A,P)*distant(B,P)));
}

/*凸包判断*/
bool IsConvexBag(pos* Vectex)
{
	int direction=0;
	//保存点集Vectex的旋转方向direction   1:右手正螺旋，逆时针   -1:左手正螺旋，顺时针
	for(int i=0;i<=n-1;i++)
	{
		int temp=precision(cross(Vectex[i],Vectex[i+1],Vectex[i+1],Vectex[i+2]));

		if(!direction)   //避免最初的点出现共线的情况
			direction=temp;

		if(direction*temp<0)  //只要Vectex是凸包，那么无论Vectex的旋转方向如何，direction*temp都会>=0
			return false;
	}
	return true;
}

/*判断点与多边形的关系*/
bool IsIn(pos* Vectex)
{
	double CircleAngle=0.0;  //环绕角
	for(int i=1;i<=n;i++)  //注意重复边不计算
		if(precision(cross(Peg,Vectex[i],Peg,Vectex[i+1]))>=0)
			CircleAngle+=angle(Vectex[i],Vectex[i+1],Peg);
		else
			CircleAngle-=angle(Vectex[i],Vectex[i+1],Peg);

	if(precision(CircleAngle)==0)  //CircleAngle=0, Peg在多边形外部
		return false;
	else if(precision(CircleAngle-pi)==0 || precision(CircleAngle+pi)==0)   //CircleAngle=180, Peg在多边形边上(不包括顶点)
	{
		if(precision(PegR)==0)
			return true;
	}
	else if(precision(CircleAngle-2*pi)==0 || precision(CircleAngle+2*pi)==0)   //CircleAngle=360, Peg在多边形边内部
		return true;
	else   //CircleAngle=(0,360)之间的任意角， Peg在多边形顶点上
	{
		if(precision(PegR)==0)
			return true;
	}
	return false;
}

/*判断圆与多边形的关系*/
bool IsFit(pos* Vectex)
{
	for(int i=0;i<=n;i++)
	{
		int k=precision(fabs(cross(Peg,Vectex[i],Peg,Vectex[i+1])/distant(Vectex[i],Vectex[i+1]))-PegR);
		if(k<0)
			return false;
	}
	
	return true;
}
Posted Updated  Semprathlon / Simfae Dean ACM-ICPC / Programing4 minutes read (About 596 words)

POJ 1556 The Doors（线段交+最短路） - kuangbin - 博客园

POJ 1556 The Doors（线段交+最短路） - kuangbin - 博客园.

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/************************************************************
 * Author        : kuangbin
 * Email         : kuangbin2009@126.com 
 * Last modified : 2013-07-14 10:47
 * Filename      : POJ1556TheDoors.cpp
 * Description   : 
 * *********************************************************/

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>

using namespace std;

const double eps = 1e-8;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0) return -1;
    else return 1;
}
struct Point
{
    double x,y;
    Point(){}
    Point(double _x,double _y)
    {
        x = _x;y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};
struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e)
    {
        s = _s;e = _e;
    }
};
//判断线段相交
bool inter(Line l1,Line l2)
{
    return 
        max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
        max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
        max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
        max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
        sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s)) <= 0 &&
        sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s)) <= 0;
}
double dist(Point a,Point b)
{
    return sqrt((b-a)*(b-a));
}
const int MAXN = 100;
Line line[MAXN];
double dis[MAXN][MAXN];
const double INF = 1e20;
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    double x,y1,y2,y3,y4;
    while(scanf("%d",&n) == 1)
    {
        if(n == -1) break;
        for(int i = 1;i <= n;i++)
        {
            scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
            line[2*i-1] = Line(Point(x,y1),Point(x,y2));
            line[2*i] = Line(Point(x,y3),Point(x,y4));
        }
        for(int i = 0;i <= 4*n+1;i++)
            for(int j = 0;j <= 4*n+1;j++)
            {
                if(i == j)dis[i][j] = 0;
                else dis[i][j] = INF;
            }
        for(int i = 1;i <= 4*n;i++)
        {
            int lid = (i+3)/4;
            bool flag = true;
            Point tmp;
            if(i&1)tmp = line[(i+1)/2].s;
            else tmp = line[(i+1)/2].e;
            for(int j = 1;j < lid;j++)
                if(inter(line[2*j-1],Line(Point(0,5),tmp)) == false
                        && inter(line[2*j],Line(Point(0,5),tmp)) == false)
                    flag = false;
            if(flag)dis[0][i] =dis[i][0] = dist(Point(0,5),tmp);
            flag = true;
            for(int j = lid+1;j <= n;j++)
                if(inter(line[2*j-1],Line(Point(10,5),tmp)) == false
                        && inter(line[2*j],Line(Point(10,5),tmp)) == false)
                    flag = false;
            if(flag)dis[i][4*n+1] =dis[4*n+1][i] = dist(Point(10,5),tmp);
        }
        for(int i = 1;i <= 4*n;i++)
            for(int j = i+1;j <=4*n;j++)
            {
                int lid1 = (i+3)/4;
                int lid2 = (j+3)/4;
                bool flag = true;
                Point p1,p2;
                if(i&1)p1 = line[(i+1)/2].s;
                else p1 = line[(i+1)/2].e;
                if(j&1)p2 = line[(j+1)/2].s;
                else p2 = line[(j+1)/2].e;
                for(int k = lid1+1;k < lid2;k++)
                    if(inter(line[2*k-1],Line(p1,p2)) == false
                            && inter(line[2*k],Line(p1,p2)) == false)
                        flag = false;
                if(flag) dis[i][j] = dis[j][i] = dist(p1,p2);
            }
        bool flag = true;
        for(int i = 1;i <= n;i++)
            if(inter(line[2*i-1],Line(Point(0,5),Point(10,5))) == false
                    && inter(line[2*i],Line(Point(0,5),Point(10,5))) == false)
                flag = false;
        if(flag)dis[0][4*n+1] = dis[4*n+1][0] = 10;
        for(int k = 0;k <= 4*n+1;k++)
            for(int i = 0;i <= 4*n+1;i++)
                for(int j = 0;j <= 4*n+1;j++)
                    if(dis[i][k] + dis[k][j] < dis[i][j])
                        dis[i][j] = dis[i][k] + dis[k][j];
        printf("%.2lf\n",dis[0][4*n+1]);
    }
    
    return 0;
}
Posted Updated  Semprathlon / Simfae Dean a minute read (About 117 words)

My Git Note

Note! 益rz

Unix的哲学是“没有消息就是好消息”

基本操作

-新增文件：git add {readme.txt}

-提交并注释：git commit -m “{wrote a readme file}”

-仓库初始化：git init

-取得仓库状态：git status

分支管理

参考指针或链表模型

-新建分支并切换：git checkout [-b] {dev}

-查看分支：git branch

-切换分支：git checkout [master]

-合并分支：git merge

-删除分支：git branch -d

Posted Updated  Semprathlon / Simfae Dean Deployment4 minutes read (About 552 words)

使用github+jekyll搭建blog环境，完美替代wordpress-前端随记

使用github+jekyll搭建blog环境，完美替代wordpress-前端随记-黑妞哈哈.

也来讲讲怎么使用github pages做blog环境

优点

空间免费，github托管，稳定又安全，遭遇过空间商跑路的朋友是不是想起伤心往事；

允许本地服务器调试，脱离网络写文章毫无压力，因为可以使用git命令同步来管理文章，版本控制妥妥的，对技术人员来说，一键恢复，实在是神物；

还能绑定顶级域名，亲，人家免费空间竟然还允许我们绑域名有木有~~；

文章用markedown编写，以前遭受排版困扰的亲们是不是很激动；

购买域名

（略）

用免费的dnsPod做域名解析

dnspod链接地址https://www.dnspod.cn/ dnspod settings

github注册和本地电脑jekyll等环境配置

参考最底下的参考文章，省略。。。

命令

  1. git命令获取远程文件

    git clone git@github.com:heiniuhaha/heiniuhaha.github.com.git

  2. 定位到目录heiniu.github.com

    cd .ssh/heiniuhaha.github.com

  3. 使用rake命令

    rake page           # Create a new page.
rake post           # Begin a new post in ./_posts
rake preview        # Launch preview environment

  4. 写文章的时候学习下markdown语法

    如:中文单引号 ` 用来标注小块代码,如github jekyll

  5. 最后提交git代码

    git add .
git commit . -m 'just another commit'

日常发布完整命令

git clone git@github.com:heiniuhaha/heiniuhaha.github.com.git//本地如果无远程代码，先做这步，不然就忽略
cd .ssh/heiniuhaha.github.com//定位到你blog的目录下
git pull origin master //先同步远程文件，后面的参数会自动连接你远程的文件
git status //查看本地自己修改了多少文件
git add .//添加远程不存在的git文件
git commit * -m "what I want told to someone"
git push origin master //更新到远程服务器上

参考文章

附件：git api 总结图

git-api
Posted Updated  Semprathlon / Simfae Dean a few seconds read (About 0 words)

Guide

Posted Updated  Semprathlon / Simfae Dean a few seconds read (About 0 words)

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