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…

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…

#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…

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 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…

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…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=3549 同样的网络最大流 T了好几次原因是用了cout,改成printf就A了 还有HDU oj的编译器也不支持以下的写法 G[from].push_back((edge){to,cap,G[to].size()}); G[to].push_back((edge){from,0,G[from].size() - 1}); #include<iostream> #include<cstdio> #i…

http://acm.hdu.edu.cn/showproblem.php?pid=3549 刚接触网络流,感觉有点难啊,只好先拿几道基础的模板题来练练手. 最大流的模板题. #include<iostream> #include<cstring> #include<string> #include<algorithm> #include<queue> using namespace std; int n, m, flow; ]; //路径记录…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=3549 Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 5111    Accepted Submission(s): 2385 Problem Description Network flow is a well-kn…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 Problem Description 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.   Input The first line of inp…

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…

//////////在这幅图中我们首先要增广1->2->4->6,这时可以获得一个容量为2的流,但是如果不建立4->2反向弧的话,则无法进一步增广,最终答案为2,显然是不对的,然而如果建立了反向弧4->2,则第二次能进行1->3->4->2->5->6的增广,最大流为3. #include<stdio.h> #include<algorithm> #include<string.h> #include<q…

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];…

Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 8199    Accepted Submission(s): 3814 Problem Description Network flow is a well-known difficult problem for ACMers. Given a graph, yo…

Problem Description 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. Input The first line of input contains an integer T, denoting the number of test…

Ford-Fulkerson方法依赖于三种重要思想,这三个思想就是: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流,从而增加流的值.反复进行这一过程,直至增广路 径都被找出来,根据最大流最小割定理,当不包含增广路径时,f是G中的一个最大流. #include <stdio…

题意:给个赤裸的最大流问题. 思路:EK+BFS解决.跟HDU1532几乎一样的. #include <bits/stdc++.h> #define LL long long #define pii pair<int,int> #define INF 0x7f7f7f7f using namespace std; ; int cap[N][N]; int flow[N][N]; int a[N]; int path[N]; vector<int> vect[N]; in…

Flow Problem 题意:N个顶点M条边,(2 <= N <= 15, 0 <= M <= 1000)问从1到N的最大流量为多少? 分析:直接使用Edmonds_Karp算法即可:下面是对增广路的一些理解和代码的解释: 残量:容量-流量: 增广:求出从源点到汇点的一条道路中所有残量的最小值d,把对应的所有边上的流量增加d,反向边(t->s)流量减少d(反向边的cap其实一直是0,只是flow为负了); 技巧:这次的ins的标号是从0开始的,即tot++,之前我都是++t…

Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 15345    Accepted Submission(s): 7234 Problem Description Network flow is a well-known difficult problem for ACMers. Given a graph,…

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…

其实还是不是很懂dinic----- 抄了一个模板--- http://www.cnblogs.com/naturepengchen/articles/4403408.html 先放在这里--- #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; ; << ) - ;…

题目链接 Problem Description 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. Input The first line of input contains an integer T, denoting the number of…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3549 Flow Problem Description 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. Input The first line of…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=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. Input The first line of input contains an integer…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=3549 模板题,白书上的代码... #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; <<; ][],flow[][],n; int Edmonds_Karp(in…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 题意: 给你一个有向图,问你1到n的最大流. dinic模版 (n*n*m) #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; //点个数 ; //边个数 const int inf = 0x3f3f3f3f; struct Edge { int to, ne…

大致题意:     给出一个又n个点,m条边组成的无向图.给出两个点s,t.对于图中的每个点,去掉这个点都需要一定的花费.求至少多少花费才能使得s和t之间不连通. 大致思路:     最基础的拆点最大流,把每个点拆作两个点 i 和 i' 连接i->i'费用为去掉这个点的花费,如果原图中有一条边a->b则连接a'->b.对这个图求出最大流即可. 画了个图,仔细看看似乎是这么回事 /* HDU 4289 G++ 62ms 1888K 最大流 SAP */ #include<stdio.…

http://acm.hdu.edu.cn/showproblem.php?pid=1533 给一个网格图,每两个点之间的匹配花费为其曼哈顿距离,问给每个的"$m$"匹配到一个"$H$"的最小花费 思路: 实际上应该是一道KM匹配的题,不过也可以转化为费用流 建立附加源汇点,$m$连接源点,$H$连接汇点,其余建立$num(H)*num(m)$条边,容量为1,费用为曼哈顿距离,跑一边费用流就行.... #include <bits/stdc++.h> #…