poj 1459 多源汇网络流 ISAP】的更多相关文章

题意: 给n个点,m条边,有np个源点,nc个汇点,求最大流 思路: 超级源点把全部源点连起来.边权是该源点的最大同意值: 全部汇点和超级汇点连接起来,边权是该汇点的最大同意值. 跑最大流 code: #include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<vector> #include<string> #include&l…
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…
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…
题意:在一图中,每个点有俩个属性:现在牛的数量和雨棚大小(下雨时能容纳牛的数量),每个点之间有距离, 给出牛(速度一样)在顶点之间移动所需时间,问最少时间内所有牛都能避雨. 模型分析:多源点去多汇点(此题源点也可能是汇点(源汇同点)),所以我的方法是:建立超级源点和超级 汇点,超级源点连想所有点,容量为该店本来的牛数量,在把各点拆成如图(略丑),到汇点的容量分别为 避雨容纳量,点点之间容量可以无限大.folyd求任意两点的最短路后,二分最大时间建图,枚举求最大之最小即可. 未1A原因: 1:开始…
题意:给一个矩阵,给出每行每列之和,附加一些条件,如第i行第j列数必需大于(小于)多少. 思路题解:矩阵模型,模拟网络流,行.列标号为结点,构图,附加s,t,s连行标(容量上下限每行之和(必需以这个 值全部送过去),每个列标连向t(容量上下限每列之和)),其他每个行到列都有边(有限制的按限制 来,无限制的自己添加).由于s->网络->t,若有解,总流量f必为 矩阵所有数之和,故添加边t-S,容量上下线为矩阵和sum,这样保证了每个点出入流相等,满足流量平衡, 转化为无源无汇有上下线可行流问题,…
Wolsey“强整数规划模型”经典案例之一单源固定费用网络流问题 阅读本文可以理解什么是“强”整数规划模型. 单源固定费用网络流问题见文献[1]第13.4.1节(p229-231),是"强整数规划建模“的极好案例. 本文是本博客原创,本博客不转贴他人作品. 单源固定费用网络流问题(The Signle Source Fixed Charge Network Flow Problem) 单源固定费用网络流问题:给定一个有向网络 (边数为m,节点数n),网络上只有一个流量流入节点(标记为1),但有若…
POJ 2135 Farm Tour (网络流,最小费用最大流) Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the b…
POJ 2516 Minimum Cost (网络流,最小费用流) Description Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marke…
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…
Description Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others. Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although…
Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to cr…
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…
Description We are supposed to make a budget proposal for this multi-site competition. The budget proposal is a matrix where the rows represent different kinds of expenses and the columns represent different sites. We had a meeting about this, some t…
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…
题意: 2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20 2 1 1 2 表示 共有2个节点,生产能量的点1个,消耗能量的点1个, 传递能量的通道2条:(0,1)20 (1,0)10 代表(起点,终点)最大传递的能量 (0)15 (产生能量的点)产生的最大能量(1)20 (消费能量的点)消费的最大能量 初学网络流,我想从基础练起:就先用EK算法写一遍 这道题看似很难,但其实只要加一个源点以及汇点,让所有的产生能量的点指向源点,让所有的消费能量的点指向汇点: #include…
题目链接 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) <= pmax(u) of power, ma…
Dining Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24945   Accepted: 10985 Description Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others. Farmer John has cooked fabul…
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: 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<cstring> #include<cstdio> #define FOR(i,f_start,f_end) for(int i=f_startl;i<=f_end;i++) #define MS(arr,arr_value) memset(arr,arr_value,sizeof(arr)) ; ; int size; int n; const int inf=0x3f3f3f3f; using namespace std; int head[maxn];…
Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 24930   Accepted: 12986 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: 26688   Accepted: 13874 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied…
题目链接 本篇博客延续上篇博客(最大流Dinic算法)的内容,此次使用EK算法解决最大流问题. EK算法思想:在图中搜索一条从源点到汇点的扩展路,需要记录这条路径,将这条路径的最大可行流量 liu 增加到结果ans中,然后反向从汇点到源点更新这条路径上的每条边的权值(减去此次的liu),同时反向边的权值也需要更新(加上此次的liu).然后再搜索新的扩展路……,循环,直到找不到新的扩展路,此时的ans就是最大流了. 注:EK算法解决最大流时,我看别人都是使用矩阵建立的图,这样反向更新扩展路径上的边…
Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and sin…
Problem Description Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory…
#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…
http://poj.org/problem?id=1459 题意:有np个发电站,nc个消费者,m条边,边有容量限制,发电站有产能上限,消费者有需求上限问最大流量. 思路:S和发电站相连,边权是产能上限,消费者和T相连,边权是需求上限,边的话就按题意加就好了.难点更觉得在于输入..加个空格..边数组要*2,因为有反向边. #include <cstdio> #include <algorithm> #include <iostream> #include <cs…
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…
题意 (混合图的欧拉回路判定) 给你一个既存在有向边, 又存在无向边的图. 问是否存在欧拉回路. \(N ≤ 200, M ≤ 1000\) 题解 难点在于无向边. 考虑每个点的度数限制. 我们先对无向边任意定向, 现在每个点都有一个出度和入度的差; 而我们要求最终每个点出度和入度相等. 令它出度减去入度为 \(deg​\) ,如果 \(deg​\) 为奇数那么必不存在欧拉回路,因为每次我们修改一条边的定向,会使得入度 \(+1​\) 出度 \(-1​\) (或者相反).那么变化后的 \(deg…
两道比较基础的网络流题目,重点就是建图. 1458:题意就是给你一些东西它们的数据,其中一些是发电站,还有一些是用户的家里,其中还有一些是中转站.让你求最大的输送电量. 就是一道很基础的最大流题目,建超级源和汇,分别向发电站连边,从用户那连进边. 具体的读入只需要写一下读优即可去括号.. CODE #include<cstdio> #include<cstring> using namespace std; ,INF=1e9; struct edge { int to,next,c…