http://poj.org/problem? id=1815 Friendship Time Limit: 2000MS   Memory Limit: 20000K Total Submissions: 9026   Accepted: 2534 Description In modern society, each person has his own friends. Since all the people are very busy, they communicate with ea…

Friendship Time Limit: 2000MS   Memory Limit: 20000K Total Submissions: 10744   Accepted: 2984 Description In modern society, each person has his own friends. Since all the people are very busy, they communicate with each other only by phone. You can…

Friendship Time Limit: 2000MS   Memory Limit: 20000K Total Submissions: 10626   Accepted: 2949 Description In modern society, each person has his own friends. Since all the people are very busy, they communicate with each other only by phone. You can…

题目链接:http://poj.org/problem?id=1815 In modern society, each person has his own friends. Since all the people are very busy, they communicate with each other only by phone. You can assume that people A can keep in touch with people B, only if 1. A kno…

Friendship Time Limit: 2000MS   Memory Limit: 20000K Total Submissions: 8025   Accepted: 2224 Description In modern society, each person has his own friends. Since all the people are very busy, they communicate with each other only by phone. You can…

妖怪题目,做到现在:2017/8/19 - 1:41…… 不过想想还是值得的,至少邻接矩阵型的Dinic算法模板get√ 题目链接:http://poj.org/problem?id=1815 Time Limit: 2000MS Memory Limit: 20000K Description In modern society, each person has his own friends. Since all the people are very busy, they communic…

http://poj.org/problem?id=1815 题意: 在现代社会,每个人都有自己的朋友.由于每个人都很忙,他们只通过电话联系.你可以假定A可以和B保持联系,当且仅当:①A知道B的电话号码:②A知道C的电话号码,而C能联系上B.如果A知道B的电话号码,则B也知道A的电话号码. 思路:这题是要我们删点,既然是删点,那么就要拆点,容量就是1. 接下来凡是能联系的,就连边,容量为INF,因为我们不是要删除这些边.跑遍最大流就能算出至少要删除多少个点. 这道题的关键是要字典序顺序输出最小割…

Friendship Time Limit: 2000MS   Memory Limit: 20000K Total Submissions: 11429   Accepted: 3173 Description In modern society, each person has his own friends. Since all the people are very busy, they communicate with each other only by phone. You can…

[题意]给出一个无向图,和图中的两个点s,t.求至少去掉几个点后才能使得s和t不连通,输出这样的点集并使其字典序最大. 不错的题,有助于更好的理解最小割和求解最小割的方法~ [思路] 问题模型很简单,就是无向图的点连通度,也就是最小点割集.麻烦之处在于需要使得点割集方案的字典序最大.这样的话通常的dfs划分点集的方法就行不通了.我们采取贪心的策略:枚举1~N的点,在残留网络中dfs检查其代表的边的两端点是否连通,如果不连通则该点可以为点割,那么我们就让他是点割,加入到答案中,然后删掉这条边更新最…

题意: 就在一个给定的无向图中至少应该去掉几个顶点才干使得s和t不联通. 算法: 假设s和t直接相连输出no answer. 把每一个点拆成两个点v和v'',这两个点之间连一条权值为1的边(残余容量) v和v''各自是一个流进的点.一个流出的点. 依据求最小割的性质.权值小的边是可能被选择的(断开的). 加入源点st=0和汇点en=2*n+1,源点与s连权值为inf的边.t''与汇点连权值为inf的边. s与s'',t与t''连权值为inf的边,这样保证自己和自己是不会失去联系的. 假设i和j有…