LINK 经典傻逼套路 就是把所有边双缩点之后叶子节点的个数 //Author: dream_maker #include<bits/stdc++.h> using namespace std; //---------------------------------------------- //typename typedef long long ll; //convenient for #define fu(a, b, c) for (int a = b; a <= c; ++a)…
首先来分析一下,这是一张无向图,要求没有两条路联通的点对个数 有两条路连通,无向图,也就是说,问题转化为不在一个点双连通分量里的点对个数 tarjan即可,和求scc还不太一样-- #include<iostream> #include<cstdio> using namespace std; const int N=5005; int n,m,h[N],cnt=1,x,y,dfn[N],low[N],tmp,s[N],top,bl[N],col,d[N],ans; bool v[…
1718: [Usaco2006 Jan] Redundant Paths 分离的路径 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 964 Solved: 503[Submit][Status][Discuss] Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another f…
Redundant Paths 分离的路径 题目描述 In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of o…