hdu 4185 Oil Skimming 二分图匹配
1 | /** Aug 24, 2015 11:15:57 AM |
折腾了过长的时间,原因:
这不是通常意义上的可随意匹配的二分图。
既然是二分图,就有把节点分为两类的依据。
由题中给定的结合规则,一定是一个(x+y)为偶数的点与一个(x+y)为奇数的点相匹配,且是相邻点的匹配。
所以会将两类点分别加入两个表中,并进行hash操作节约存储空间。
“奇”的点总是不少于“偶”的点?
1 | /** Aug 24, 2015 11:15:57 AM |
折腾了过长的时间,原因:
这不是通常意义上的可随意匹配的二分图。
既然是二分图,就有把节点分为两类的依据。
由题中给定的结合规则,一定是一个(x+y)为偶数的点与一个(x+y)为奇数的点相匹配,且是相邻点的匹配。
所以会将两类点分别加入两个表中,并进行hash操作节约存储空间。
“奇”的点总是不少于“偶”的点?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Given a string S and two integers L and M, we consider a substring of S as “recoverable” if and only if
(i) It is of length M*L;
(ii) It can be constructed by concatenating M “diversified” substrings of S, where each of these substrings has length L; two strings are considered as “diversified” if they don’t have the same character for every position.
Two substrings of S are considered as “different” if they are cut from different part of S. For example, string “aa” has 3 different substrings “aa”, “a” and “a”.
Your task is to calculate the number of different “recoverable” substrings of S.
The input contains multiple test cases, proceeding to the End of File.
The first line of each test case has two space-separated integers M and L.
The second ine of each test case has a string S, which consists of only lowercase letters.
The length of S is not larger than 10^5, and 1 ≤ M * L ≤ the length of S.
For each test case, output the answer in a single line.
3 3
abcabcbcaabc
2
2013 Asia Regional Changchun
纯粹是寻找一个最佳的hash姿势?而且还要懂得合适的时间优化?
真的TLE了好几次,现场赛的难点
枚举字串起点不必从头找到尾,因为向后滚的操作涵盖了i>=l以后的字串
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