Lint Code 1365. Minimum Cycle Section】的更多相关文章

这题可以看作POJ 1961 最小循环节的一个简化版本.某补习班广告贴里给出的两个指针的参考解法简直大误. 受POJ 1961的启发,把数组看作字串,观察可知,如果字串全部由循环节构成(包括最后一段是不完整循环节的情况),则字串刨去最后一个字符的最长匹配前缀为最小循环节.而“最后一个字符的最长匹配前缀”即为KPM里的pattern[length -1].所以我们只需按标准KMP算法求一遍失配函数即可. int minimumCycleSection(vector<int> &a) {…
题目链接:http://www.lintcode.com/zh-cn/problem/max-points-on-a-line/# 条件:给一个点数组 目标:求出共线的点的最多个数 实现:时间复杂度--O(n^2) 要考虑的特殊情况是:①有相同点(这个也太特喵隐蔽了)②斜率不存在的点 思路:暴力求解,遍历每一个点,与他之后的点进行匹配,用一个map<double,int>存储斜率对应的个数. 代码: /** * Definition for a point. * struct Point {…
文件链接 Karp在1977年的论文,讲述了一种\(O(nm)\)的算法,用来求有向强连通图中最小平均权值回路(具体问题请参照这里) 本人翻译(有删改): 首先任取一个节点 \(s\) ,定义 \(F_k(v)\) 为从 \(s\) 到 \(v\) 恰好经过 \(k\) 条边的最短路(不存在则为 \(\infty\) ), \(\lambda^*\) 表示答案,则 Theorem 1 \[\tag{1}\label{theorem}\lambda^* = \min_{v \in V} \max_…
类名注释报黄: 去掉这段黄做法:alt+enter 本文参考: http://www.cnblogs.com/xxoome/p/6677170.html…
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