Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16314    Accepted Submission(s): 7748 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie'…

题解: 双倍经验 dinic模板题 然后ff过不了... 代码: #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; ; int q[N*N],n,m,sum,dis[N],x,y,ans,f[N],z,a[N][N]; int bfs() { memset(dis,0xff,sizeof dis); dis[]=; ,r=;…

1.TreeMap和TreeSet类:A - Language of FatMouse ZOJ1109B - For Fans of Statistics URAL 1613 C - Hardwood Species POJ 2418D - StationE - Web Navigation ZOJ 1061F - Argus ZOJ 2212G - Plug-in2.SegmentTreeA-敌兵布阵 hdu 1166B - I Hate It HDU 1754C - A Simple Pro…

参照<挑战程序设计竞赛> #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; ; const int INF=0x7fffffff; int n,m;//n:edges,m:…

Dinic算法: 层次图:根据源点到该点的距离建图,这里设相邻的都差1. (下面部分转) 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. 请仔细阅读:在上面的例子中,我们可以看出,最终结果是1->2->5->6和1->2…

看着这个博客 然后敲了hdu1532这个入门题,算是对最大流有点理解了 #include <stdio.h> #include <string.h> #include <queue> using namespace std; <<; + ; int n,m; int cap[N][N],flow[N],pre[N]; int bfs(int src, int des)//每次用bfs找到最小的delta { queue<int> q; ;i<…

Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 76000   Accepted: 29530 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…

模板题:POJ1273 EK: #include <queue> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int n,m,jy,map[305][305],c[305][305],a[305],p[305]; bool vis[305]; int M(int e) { int f=0; queue<int>q; while(1){…

网络流一直没学,来学一波网络流. https://vjudge.net/problem/POJ-1273 题意:给定点数,边数,源点,汇点,每条边容量,求最大流. 解法:EK或dinic. EK:每次增广用bfs选择一条从源到汇具有最少边数的增广路径,然后找出该路径容量最小的边,就是此次增加的流量,然后沿该路径增加反向边,同时修改每条边的容量,重复上述过程直到找不到增广路(即minFlow = 0)为止. dinic: 每次bfs从源点到汇点分层(层数是源点到它最少要经过的边数),然后dfs从源…

HDU-1532 题意简单的来说就是从1点到n点,最大的流量是多少. 代码: #include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #def…