题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1853 There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city b…

描述 When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000)…

Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18150   Accepted: 7023 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of…

题目链接:http://poj.org/problem?id=2135 Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17672   Accepted: 6851 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000…

Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17230   Accepted: 6647 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of…

题目链接 题意:无向图有N(N <= 1000)个节点,M(M <= 10000)条边:从节点1走到节点N再从N走回来,图中不能走同一条边,且图中可能出现重边,问最短距离之和为多少? 思路:很经典的构图(看题解的);每条原图中的边赋予cap为1,表示只走一次.超级源点s和汇点t分别和起点终点连边,cap为2,这里cap为2就直接限制了只能有两次最大流:同时最大流中以权值限制得到的就是最小费用:很注意的一点就是此题为无向图带权值,建图时每条有向边建成两条即总边数为4*M.由于spfa找最短路是有…

题意: 有n个点和m条边,让你从1出发到n再从n回到1,不要求所有点都要经过,但是每条边只能走一次.边是无向边. 问最短的行走距离多少. 一开始看这题还没搞费用流,后来搞了搞再回来看,想了想建图不是很难,因为要保证每条边只能走一次,那么我们把边拆为两个点,一个起点和终点,容量是1,权重是这条路的长度.然后两个端点分别向起点连接容量是1权重是0的边,终点分别向两个端点连容量是1权重是0的边,从源点到1连容量为2权重为0的边,从n到汇点连容量为2权重为0的边. #include<stdio.h>…

原题 费用流板子题. 费用流与最大流的区别就是把bfs改为spfa,dfs时把按deep搜索改成按最短路搜索即可 #include<cstdio> #include<queue> #include<cstring> #define N 20020 using namespace std; int n,m,src, des, head[N],dis[N],cur[N],ans,cnt=2,s,t, ANS; queue <int> q; bool vis[N]…

题目大概说给一张有向图,每条边都有权值,要选若干条边使其形成若干个环且图上各个点都属于且只属于其中一个环,问选的边的最少权值和是多少. 各点出度=入度=1的图是若干个环,考虑用最小费用最大流: 每个点拆成两点u和u' 源点向u连容量1费用0的边,表示这个点的出度最多为1 u'向汇点连容量1费用0的边,表示这个点的入度最多为1 对于原图上任何一条有向边<a,b,c>,a向b'连容量1费用c的边,选择这条边后a的出度+1,b的入度+1,费用+c 这样跑最小费用最大流,如果没满流就无解,否则MCMF…

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 1197    Accepted Submission(s): 626 Problem Description There are N cities in our country, and M one-way roads connecting them. Now Li…