poj2125Destroying The Graph(最小割+输出方案)

【poj2125Destroying The Graph(最小割+输出方案)】的更多相关文章

poj2125Destroying The Graph(最小割+输出方案)

题目请戳这里题目大意:给一张有向图,现在要选择一些点,删掉图中的所有边.具体操作为:选择点i,可以选择删除从i出发的所有有向边或者进入i的所有有向边,分别有个代价ini和outi,求最小的代价删掉所有边.并输出删除方案. 题目分析:最小割.因为每次删除的是一个点的所有入边或者所有出边.那么就很明显了,拆点,将i拆成i和I+n2个点,分别表示第i个点的入度点和出度点.源点到每个i连边,表示从i点的出边,边权为outi,i+n表示第i个点的入度点,那么i+n-->汇点建边,边权为ini,对于每对给…

poj 2125 Destroying The Graph 最小割+方案输出

构图思路: 1.将所有顶点v拆成两个点, v1,v2 2.源点S与v1连边,容量为 W- 3.v2与汇点连边,容量为 W+ 4.对图中原边( a, b ), 连边 (a1,b2),容量为正无穷大则该图的最小割(最大流)即为最小花费. 简单证明: 根据ST割集的定义,将顶点分成两个点集.所以对于原图中的边(a,b),转换成 S->a1->b2->T. 则此时路径必定存在一条割边,因为a1->b2为无穷大,所以割边必定是 S->a1 or b2->T, 若为前者则意味…

CodeForces1082G Petya and Graph 最小割

网络流裸题 \(s\)向点连边\((s, i, a[i])\) 给每个边建一个点边\((u, v, w)\)抽象成\((u, E, inf)\)和\((v, E, inf)\)以及边\((E, t, w)\) 最小割建模... 然后就没了....复习一下板子吧 #include <cstdio> #include <cstring> #include <iostream> using namespace std; #define ll long long #defin…

POJ 2125 Destroying The Graph [最小割打印方案]

Destroying The Graph Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8311 Accepted: 2677 Special Judge Description Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that B…

Petya and Graph(最小割,最大权闭合子图)

Petya and Graph http://codeforces.com/contest/1082/problem/G time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Petya has a simple graph (that is, a graph without loops or multiple edges) con…

POJ2125 Destroying The Graph (最小点权覆盖集）（网络流最小割）

Destroying The Graph Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8158 Accepted: 2620 Special Judge Description Alice and Bob play the following game. First, Alice draws some d…

Petya and Graph/最大权闭合子图、最小割

原题地址:https://codeforces.com/contest/1082/problem/G G. Petya and Graph time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Petya has a simple graph (that is, a graph without loops or multiple e…