题目链接: http://poj.org/problem?id=1459 因为发电站有多个,所以需要一个超级源点,消费者有多个,需要一个超级汇点,这样超级源点到发电站的权值就是发电站的容量,也就是题目中的pmax,消费者到超级汇点的权值就是消费者的容量,也就是题目中的cmax.初学网络流,第一眼看到这个题还以为应该先做一遍EK算法,然后减去max(p-pmax, c-cmax)呢..没想到这个题的难点就是建图而已.. #include <stdio.h> #include <string…

POJ 1459 Power Network / HIT 1228 Power Network / UVAlive 2760 Power Network / ZOJ 1734 Power Network / FZU 1161 (网络流,最大流) Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A…

题目连接 http://poj.org/problem?id=1459 Power Network Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an…

Power Network Time Limit: 2000MS Memory Limit: 32768K Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produc…

Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 25514   Accepted: 13287 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied…

Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 22987   Accepted: 12039 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied…

#include<cstdio> #include<cstring> #include<algorithm> #include<queue> #include<vector> #define INF 1e9 using namespace std; const int maxn=100+5; struct Edge { int from,to,cap,flow; Edge(){} Edge(int f,int t,int c,int fl):fr…

点击打开链接 Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 20903   Accepted: 10960 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be su…

1.看了好久,囧. n个节点,np个源点,nc个汇点,m条边(对应代码中即节点u 到节点v 的最大流量为z) 求所有汇点的最大流. 2.多个源点,多个汇点的最大流. 建立一个超级源点.一个超级汇点,然后求超级源点到超级汇点的最大流即可. 3. 1.SAP邻接矩阵形式: /* SAP算法(矩阵形式) 结点编号从0开始 */ #include<iostream> #include<stdio.h> #include<string.h> using namespace std…

题意:给出n,np,nc,m,n为节点数,np为发电站数,nc为用电厂数,m为边的个数.      接下来给出m个数据(u,v)z,表示w(u,v)允许传输的最大电力为z:np个数据(u)z,表示发电站的序号,以及最大的发电量:      nc个数据(u)z,表示用电厂的序号,以及最大的用电量.      最后让你求可以供整个网络使用的最大电力.思路:纯模板题.      这里主要是设一个源点s和一个汇点t,s与所有发电厂相连,边的最大容量为对应发电厂的最大发电量:      t与所有用电厂相连…