src:源点 sink:汇点 #include<queue> #include<iostream> #include<string.h> #include<stdio.h> using namespace std; ; , maxm = ; struct Edge{ int v, f, nxt; }; int src, sink; ]; int nume; Edge e[maxm*+]; void addedge(int u, int v, int c){…

POJ 1273给出M条边,N个点,求源点1到汇点N的最大流量. 本文主要就是附上dinic的模板,供以后参考. #include <iostream> #include <stdio.h> #include <algorithm> #include <queue> #include <string.h> /* POJ 1273 dinic算法模板 边是有向的,而且存在重边,且这里重边不是取MAX,而是累加和 */ using namespace…

DINIC网络流+当前弧优化 const inf=; type rec=record s,e,w,next:longint; end; var b,bb,d,q,tb:..] of longint; a,ta:..] of rec; xz1,xz2:..,..]of longint; ch:char; n,m,i,j,k,l,st,ed,ww,top,top2,ans,nn,dd,x:longint; function min(aa,bb:longint):longint; begin if a…

http://poj.org/problem?id=1459 嗯,网络流模板...多源点多汇点的图,超级汇点连发电厂,用户连接超级汇点 Status Accepted Time 391ms Memory 1588kB Length 2166 Lang G++ Submitted 2018-05-23 14:43:22 Shared RemoteRunId 18625420 #include <iostream> #include <string.h> #include <cm…

妖怪题目,做到现在:2017/8/19 - 1:41…… 不过想想还是值得的,至少邻接矩阵型的Dinic算法模板get√ 题目链接:http://poj.org/problem?id=1815 Time Limit: 2000MS Memory Limit: 20000K Description In modern society, each person has his own friends. Since all the people are very busy, they communic…

Drainage Ditches Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 45 Accepted Submission(s): 38   Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorit…

#include<stdio.h> #include<queue> #include<string.h> using namespace std; #define inf 0x3fffffff #define N 200 struct node { int v,w,next; }bian[N*N*2],fbian[N*N*2]; int head[N],yong,tt; int deep[N];//深度保留层次图 void addedge(int u,int v,int…

Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 63339   Accepted: 24434 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…

题意: 农场主有f种食物,d种饮料,n头牛. 接下来的n行每行第一个数代表第i头牛喜欢吃的食物数量,和第i头牛喜欢喝的饮料数目. 接下来分别是喜欢的食物和饮料的编号. 求解:农场主最多能保证几头牛同时能吃到喜欢的食物和喜欢的饮料. 思路: 从源点到每种食物加流量为1的边,保证每种食物只能被吃一次. 将每头牛分为两个点,连一条流量为1的边,保证每头牛最多只能吃一份食物喝一份饮料. 将每种饮料和汇点都进行连接,同样流量为1. 将牛喜欢的食物和牛进行连边流量为1,将另一组代表牛的点和喜欢的饮料进行连边…

这里只是用来存放模板,几乎没有讲解,要看讲解网上应该很多吧…… ek bfs不停寻找增广路到找不到为止,找到终点时用pre回溯,O(VE^2) #include<cstdio> #include<iostream> #include<cstring> #include<queue> #include<algorithm> using namespace std; const int INF = 0x3f3f3f3f; ; ][], pre[],…