嘟嘟嘟 最小路径覆盖板子题. 建二分图,然后跑Dinic(因为我不会匈牙利),然后ans = n - maxflow(). 主要是发一下用链前存图的写法.(好像比vector短一点) #include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<cstring> #include<cstdlib> #include<cctype&…

Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7511   Accepted: 4471 Description 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 i…

Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6763   Accepted: 4034 Description 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 i…

Air Raid Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6528    Accepted Submission(s): 4330 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1151 Description: Consider a town where all the stree…

Air Raid Language:Default Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9547 Accepted: 5696 Description Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known t…

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 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…

Air Raid Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4591    Accepted Submission(s): 3072 Problem Description Consider a town where all the streets are one-way and each street leads from one…

Air Raid Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3378    Accepted Submission(s): 2223 Problem Description Consider a town where all the streets are one-way and each street leads from one…

http://acm.hdu.edu.cn/showproblem.php?pid=1151 Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7469   Accepted: 4451 Description Consider a town where all the streets are one-way and each street leads from one intersection to an…

最小点覆盖 Air Raid Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3002    Accepted Submission(s): 1951 Problem Description Consider a town where all the streets are one-way and each street leads f…

POJ 1422 Air Raid 题目链接 题意:给定一个有向图,在这个图上的某些点上放伞兵,能够使伞兵能够走到图上全部的点.且每一个点仅仅被一个伞兵走一次.问至少放多少伞兵 思路:二分图的最小路径覆盖,每一个点拆成两个点,然后依据有向边连边,然后答案为n - 最大匹配数 代码: #include <cstdio> #include <cstring> #include <vector> #include <algorithm> using namespa…

Air Raid Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 175 Accepted Submission(s): 133   Problem Description Consider a town where all the streets are one-way and each street leads from one inte…

Air Raid Problem Description 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…

题目: Air Raid Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 124 Accepted Submission(s): 102   Problem Description Consider a town where all the streets are one-way and each street leads from one…

Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6520   Accepted: 3877 Description 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 i…

传送门:http://poj.org/problem?id=1422 Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9303   Accepted: 5570 Description Consider a town where all the streets are one-way and each street leads from one intersection to another. It is…

Air Raid Time Limit: 1000ms Memory Limit: 10000KB This problem will be judged on PKU. Original ID: 142264-bit integer IO format: %lld      Java class name: Main   Consider a town where all the streets are one-way and each street leads from one inters…

Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8577   Accepted: 5127 Description 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 i…

有向无环图(DAG)的最小路径覆盖的模板题. 定义:在一个有向图中,找出最少的路径,使得这些路径经过了所有的点. 由题意可得这是一个有向图,而路径不能相交,于是我们知道这是无向图的不相交最小路径覆盖问题 我们把所有的点拆成两个,当两点之间有路径时,我们在u与v'之间建一条容量为1的边,利用二分图的思想,最大流数即为减去的路径数. 求法:最小路径条数 = 点数 - 分裂图最大流数 Code #include<iostream> #include<cstdio> #include<…

http://poj.org/problem?id=1422 题目大意:n个点m条有向边,每条边只能走一次,往点上放人让他们走遍所有边,问至少要多少人. —————————————————————— 网上的题解都蛮详尽的了所以不想写这篇博客. (但是强迫症) 首先这是一道最小路径覆盖问题,有结论转换成求|节点数-最大匹配|,证明看这个人的博客. 我们将每个点拆分成两个点(设左点和右点),左点为一个集合,右点为一个集合. 这样的话两个点ij中间的有向边就是i左点连j右点. #include<cst…

Description 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 intersection i.e…

题目链接:http://poj.org/problem?id=1422 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 t…

题意 给定一个有向图,在这个图上的某些点上放伞兵,可以使伞兵可以走到图上所有的点.且每个点只被一个伞兵走一次.问至少放多少伞兵. 思路 裸的最小路径覆盖. °最小路径覆盖 [路径覆盖]在一个有向图G(V, E<u,v>)中,路径覆盖就是在图中找一些路经,使之覆盖了图中的所有顶点,且任何一个顶点有且只有一条路径与之关联(如果把这些路径中的每条路径从它的起始点走到它的终点,那么恰好可以经过图中的每个顶点一次且仅一次):如果不考虑图中存在回路,那么每条路径就是一个弱连通子集． [最小路径覆盖]最小路…

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…

题意:     一个镇里所有的路都是单向路且不会组成回路. 派一些伞兵去那个镇里,要到达所有的路口,有一些或者没有伞兵可以不去那些路口,只要其他人能完成这个任务.每个在一个路口着陆了的伞兵可以沿着街去到其他路口.我们的任务是求出去执行任务的伞兵最少可以是多少个. 思路: 这个题就是个最小路径覆盖问题. 路径覆盖的定义是:在有向图中找一些路径,使之覆盖了图中的所有顶点,就是任意一个顶点都跟那些路径中的某一条相关联,且任何一个顶点有且只有一条路径与之关联,一个单独的顶点是一条路径.最小路径覆盖就是最…

题目大意: 有n个城市,m条道路,城市的道路是单向.  现在我们的伞兵要降落在城市里,然后我门的伞兵要搜索所有道路.问我们最少占领多少个城市就可以搜索所有的道路了. 我们可以沿着道路向前走到达另一个城市.   题目分析: 这道题目其实就是求最小路径覆盖 最小路径覆盖问题:用尽量少的不相交简单路径覆盖有向无环图的所有顶点. 将每个顶点分为两个,分别在X集合和Y集合中,如果存在有向边(a,b),对应在二分图中有(Xa,Yb) 最小路径数=节点数-最大匹配 简单解释: 原图的路径覆盖和新图的匹配间有对…

题意:给一个有向无环图,求出来最少需要几个士兵可以遍历所有的边. 分析:有向无环图的最小边覆盖 = 点数 - 最大匹配数 为什么是这样的公式??可以思考一下,如果这N个点之间没有边,是不是应该有N个士兵去查看,但是如果增加一条边就应该减去这条边,以此类推,公式就比较容易明白了. ******************************************************************** #include<stdio.h> #include<; ; ; j<…

很明显求最小路径覆盖 就是求最大匹配 #include <iostream> #include <cstdio> #include <cstring> #include <vector> using namespace std; ; vector<int> map[maxn]; int vis[maxn]; int link[maxn]; int n,k; bool dfs(int t) { int i,x,size=map[t].size();…