Ford-Fulkerson算法 戳戳http://www.cnblogs.com/luweiseu/archive/2012/07/14/2591573.html Ford-Fulkerson方法依赖于三种重要思想:残留网络,增广路径和割. Ford-Fulkerson方法是一种迭代的方法.开始时,对所有的u,v∈V有f(u,v)=0,即初始状态时流的值为0.在每次迭代中,可通过寻找一条"增广路 径"来增加流值.增广路径可以看成是从源点s到汇点t之间的一条路径,沿该路径可以压入更多的…

因为坑了无数次队友 要开始学习网络流了,先从基础的开始,嗯~ 这道题是最大流的模板题,用来测试模板好啦~ Edmonds_Karp模板 with 前向星 时间复杂度o(V*E^2) #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<queue> #define eps 0.000001 #define MAXN 20 #define MAX…

#include<stdio.h> #include<string.h> #include<queue> #include<iostream> using namespace std; #define inf 2000000000 #define N 30 int map[N][N]; int pre[N],flow[N],start,end; int bfs(){ int now,i; memset(pre,-1,sizeof(pre)); flow[st…

第一题:按顾客访问猪圈的顺序依次构图(顾客为结点),汇点->第一个顾客->第二个顾客->...->汇点 //第一道网络流 //Ford-Fulkerson //Time:47Ms Memory:276K #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<queue> using namespace std;…

HDU 3549 Flow Problem(最大流) Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) [Description] [题目描述] Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for t…

Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 6674    Accepted Submission(s): 3112 Problem Description Network flow is a well-known difficult problem for ACMers. Given a graph, yo…

一.Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainag…

Flow Problem HDU - 3549 Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.  InputThe first line of input contains an integer T, denoting the number of t…

网络流 HDU 3549 Flow Problem 题目:pid=3549">http://acm.hdu.edu.cn/showproblem.php?pid=3549 用增广路算法进行求解.注意的问题有两个: 1. 每次增广的时候,反向流量也要进行更行.一開始没注意,WA了几次 ORZ 2. 对于输入的数据,容量要进行累加更新. // 邻接矩阵存储 #include <bits/stdc++.h> using namespace std; const int INF = 0…

1.HDU 1532 最大流入门,n个n条边,求第1点到第m点的最大流.只用EK做了一下. #include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define F(i,a,b) for (int i=a;i<b;i++) #define FF(i,a,b) for (int i=a;i<=b;i++) #define…