http://acm.hdu.edu.cn/showproblem.php?pid=3549 Ford-Fulkerson算法. #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <string> #include <algorithm> #include <stri…

解题关键:使用的挑战程序设计竞赛上的模板,第一道网络流题目,效率比较低,且用不习惯的vector来建图. 看到网上其他人说此题有重边,需要注意下,此问题只在邻接矩阵建图时会出问题,邻接表不会存在的,也体现了邻接表的优越性? edge结构体的第三个变量为from的下标. 模板一: #include<bits/stdc++.h> #define MAX_V 17 #define inf 0x3f3f3f3f using namespace std; typedef long long ll; in…

最大流裸题,贴下模版 view code#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <vector> #include <queue> using namespace std; const int INF = 0x3f3f3f3f; const int maxn…

A Plug for UNIX Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15597   Accepted: 5308 Description You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an int…

Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 28193    Accepted Submission(s): 12476 Problem Description Network flow is a well-known difficult problem for ACMers. Given a graph,…

Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 27028    Accepted Submission(s): 11408 Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1],…

Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 17963    Accepted Submission(s): 8464 Problem Description Network flow is a well-known difficult problem for ACMers. Given a graph, y…

裸最大流,做模板用 m条边,n个点,求最大流 #include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; ; const int INF = 0x7fffffff; int flow[N]; int cap[N][N]; int pre[N]; queue<int>…

题目链接 题意 裸的最大流. 学习参考 http://www.cnblogs.com/SYCstudio/p/7260613.html Code #include <bits/stdc++.h> #define inf 0x3f3f3f3f #define maxm 1010 #define maxn 20 using namespace std; typedef long long LL; struct Edge { int to, ne, c; }edge[maxm<<1];…

题意: 1.一个人从[1,1] ->[n,n] ->[1,1] 2.仅仅能走最短路 3.走过的点不能再走 问最大和. 对每一个点拆点限流为1就可以满足3. 费用流流量为2满足1 最大费用流,先给图取负,结果再取负,满足2 #include <stdio.h> #include <string.h> #include <iostream> #include <math.h> #include <queue> #include <s…