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 Total Submissions: 25414   Accepted: 13247 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied…

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…

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…

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…

#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…

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…

题目:http://poj.org/problem?id=1459 题意:有一些发电站,消耗用户和中间线路,求最大流.. 加一个源点,再加一个汇点.. 其实,过程还是不大理解.. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; <<; ][],flow[…

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…

不可以理解的是,测评站上的0ms是怎么搞出来的. 这一题在建立超级源点和超级汇点后就变得温和可爱了.其实它本身就温和可爱.对比了能够找到的题解: (1)艾德蒙·卡普算法(2)迪尼克算法(3)改进版艾德蒙·卡普算法(MY METHOD) 不去管那个0ms的吧,那么(3)号算法最为美妙[它的别名是:ISAP],时间可观. 这个就算是一个ISAP的模板吧(除了输入的难看的几行外,其余均是标准的大米饼牌模板!) #include<stdio.h> #include<algorithm> #…

<题目链接> 题目大意:给出 n 个点,其中包括 np个发电站,nc 个消费者, 剩下的全部都是中转点,再给出 这些点中的m 条边,代表这两点间的最大传输电量,并且给出发电站的最大发送电量,以及消费者的最大承受电量,求所有消费者所能得到的最大电量. 解题分析:本题发电站可以看成源点,消费者看成汇点,由于可能有多个源点和汇点,因此我们可以建立一个超级源点和超级汇点,超级源点与所有发电站直接相连,容量为每个发电站的最大容量,超级汇点与所有消费者相连,容量为每个消费者的最大容量. #include…

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…

                                     Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 27229   Accepted: 14151 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power trans…

题目链接:http://poj.org/problem?id=1459 Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 27074   Accepted: 14066 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transp…

Language: Default Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 23407   Accepted: 12267 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node…

  Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 24788   Accepted: 12922 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 amou…

Power Network Time Limit: 2000MS Memory Limit: 32768K Total Submissions: 24867 Accepted: 12958 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with…

Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 25108   Accepted: 13077 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: 31086   Accepted: 15986 题目链接:http://poj.org/problem?id=1459 Description: A power network consists of nodes (power stations, consumers and dispatchers) connected by power trans…

Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 24019   Accepted: 12540 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied…

读题比做题难系列…… poj1087 输入n,代表插座个数,接下来分别输入n个插座,字母表示.把插座看做最大流源点,连接到一个点做最大源点,流量为1. 输入m,代表电器个数,接下来分别输入m个电器,字符串表示.把电器看做最大流终点,连接到一个点做最大汇点,流量为1. 输入k,代表转换器个数,接下来分别输入k个转换器,每个插座输入两个字母a,b表示a可以连在b上.把转换器看做流,b->a,因为转换器无限提供,流量为无限大 代码: #include <iostream> #include &…

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 amount 0 <= p(u) <= p max(u) of power, may co…

Sample Input 2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20 7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7 (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5 (0)5 (1)2 (3)2 (4)1 (5)4 7个点包括电站和用户,2个电站,3个用户,13条边,输入13条边,输入2个电站,输入3个用户 Sample Output 15 6 增加一个源点一个汇点…

题目地址:http://poj.org/problem?id=2109 /* 题意:k ^ n = p,求k 1. double + pow:因为double装得下p,k = pow (p, 1 / n); 基础知识: 类型 长度 (bit) 有效数字 绝对值范围 float 32 6~7 10^(-37) ~ 10^38 double 64 15~16 10^(-307) ~ 10^308 long double 128 18~19 10^(-4931) ~ 10 ^ 4932 2. 二分查找…

Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 20754   Accepted: 10872 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…

解题报告 这题建模实在是好建.,,好贱.., 给前向星给跪了,纯dinic的前向星居然TLE,sad.,,回头看看优化,.. 矩阵跑过了.2A,sad,,, /************************************************************************* > File Name: PowerN.cpp > Author: _nplus > Mail: jun18753370216@gmail.com > Time: 2014年07…

原题链接 题意简述 原题看了好几遍才看懂- 给出一个个点,条边的有向图.个点中有个源点,个汇点,每个源点和汇点都有流出上限和流入上限.求最大流. 题解 建一个真 · 源点和一个真 · 汇点.真 · 源点向所有原源点连容量等于其上限的边,所有原汇点向真 · 汇点连容量等于其上限的边.跑一遍最大流即可. Code //Power Network #include <cstdio> #include <cstring> #include <algorithm> using n…

点击返回:自学Zabbix之路 点击返回:自学Zabbix4.0之路 点击返回:自学zabbix集锦 自学Zabbix9.1 Network Discovery 网络发现原理 1. 网络发现简介 网络发现有什么用?网络发现怎么配置?网络发现功能让我们能更快速的部署zabbix.简化zabbix管理.并且在经常变动的环境里面也不需要花太多的精力,毕竟网络发现也能随时变化.但是此功能是无法发现网络拓扑的.基于网络的discovery指的是基于IP的discovery,即是给定zabbix一个IP范围…