传送门:hdu1151 Air Raid 题意:在一个城镇,有m个路口,和n条路,这些路都是单向的,而且路不会形成环,现在要弄一些伞兵去巡查这个城镇,伞兵只能沿着路的方向走,问最少需要多少伞兵才能把所有的路口搜一遍. 分析:有向无环图不相交最小路径覆盖数,等于节点数减去二分图的最大匹配数,对于每条弧,弧头作为X部,弧尾作为Y部.最后在求得最大匹配的基础上,没有被匹配的Y部的点就是简单路径的起点.其个数刚好就是节点数减去二分图的最大匹配数. #include <cstdio> #include…

题目链接:https://vjudge.net/problem/HDU-1151 Air Raid Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5630    Accepted Submission(s): 3785 Problem Description Consider a town where all the streets a…

Air Raid Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4959    Accepted Submission(s): 3339 Problem Description Consider a town where all the streets are one-way and each street leads from one…

Air RaidTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4974 Accepted Submission(s): 3347 Problem DescriptionConsider a town where all the streets are one-way and each street leads from one interse…

欢迎参加——BestCoder周年纪念赛(高质量题目+多重奖励) Air Raid Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3841    Accepted Submission(s): 2536 Problem Description Consider a town where all the streets are one-w…

http://acm.hdu.edu.cn/showproblem.php?pid=1151 增广路的变种2:DAG图的最小路径覆盖=定点数-最大匹配数 #include<iostream> #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> using namespace std; ; int a[N][N]; int use[N]; int matc…

刚开始看到这个题目的时候就觉得想法很明了,就是不知道如何去匹配... 去网上看了不少人的解题报告,但是对于刚接触“最小边覆盖”的我来说....还是很困难滴....于是自己又开始一如以往学习“最大独立集”.“最小点覆盖的”的思考方式啦:在了解一个看似高深的知识点之前,粗略了解这是个什么东东,那么看概念也会更好理解,(希望此博客在自己以后回头来看会一目明了,也对刚接触“最小边覆盖”的人有帮助(万分感到荣幸)): 首先把题目的第一个案例图形化(第一,图形好理解,第二,很多人看到文字就烦啦.第三,图论不…

点击打开链接 有向无环图的最小路径覆盖 = 顶点数- 最大匹配 #include <queue> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn = 150; int g[maxn][maxn]; int n, m; int link[maxn]; bool used[…

有向图的最小路径覆盖=V-二分图最大匹配. Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting from an intersection and walking through town's streets you can never reach the same inters…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1151 求最小路径覆盖 二分图最小路径覆盖=点的个数-最大匹配. 代码: #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> using namespace std; #define maxn 125 int g[maxn][maxn]; int vis[maxn]; in…