HDU 4280 Island Transport(网络流,最大流) Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships. You have a transportation company there. Some route…

HDU 4280:http://acm.hdu.edu.cn/showproblem.php?pid=4280 题意: 比较裸的最大流题目,就是这是个无向图,并且比较卡时间. 思路: 是这样的,由于是无向图,所以addedge 的反边容量直接设为原始流量.然后还可以优化搜索的方向,bfs可以从t到s跑,dfs可以从s到t跑,这样快. //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000&qu…

http://acm.hdu.edu.cn/showproblem.php?pid=4280 题意:在最西边的点走到最东边的点最大容量. 思路:ISAP模板题,Dinic过不了. #include <cstdio> #include <algorithm> #include <iostream> #include <cstring> #include <string> #include <cmath> #include <que…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4280 #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace s…

转载请注明出处:http://blog.csdn.net/u012860063 题目链接:pid=4280">http://acm.hdu.edu.cn/showproblem.php?pid=4280 Problem Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the is…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=4280 源点是West, 汇点是East, 用Dinic带入求就好了 代码:要用c++提交 #pragma comment(linker, "/STACK:102400000,102400000") ///手动开大栈区 #include<cstdio> #include<cstring> #include<iostream> #include<algo…

Island Transport Time Limit: 10000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ID: 4280 64-bit integer IO format: %I64d      Java class name: Main In the vast waters far far away, there are many islands. People are living on…

模板套起来 1 5 7 //5个结点,7个边 3 3 //坐标 3 0 3 1 0 0 4 5 1 3 3 //相连的结点和流 2 3 4 2 4 3 1 5 6 4 5 3 1 4 4 3 4 2 9 #include<cstdio> #include<algorithm> #include<cstring> using namespace std; ;//点数的最大值 ;//边数的最大值 const int INF = 0x3f3f3f3f; int n; stru…

裸的网络流,递归的dinic会爆栈,在第一行加一句就行了 #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <stdio.h> #include <string.h> #include <cstring> #include <algorithm> #include <vector> #define…

题意:有N个岛屿,M条路线,每条路都连接两个岛屿,并且每条路都有一个最大承载人数,现在想知道从最西边的岛到最东面的岛最多能有多少人过去(最西面和最东面的岛屿只有一个). 分析:可以比较明显的看出来是一个最大流的问题,而且源点汇点也都给出了,可以用最基本的dinic解决,不过注意一些优化,都则会tle的,下面是dinic AC代码 ====================================================================================…