Legal or Not HDU
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Legal or Not
HDU - 3342ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?
We all know a master can have many prentices and a prentice may have a lot of masters too, it's legal. Nevertheless,some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the same time, 3xian is HH's master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.
Please note that the "master and prentice" relation is transitive. It means that if A is B's master ans B is C's master, then A is C's master.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
If it is legal, output "YES", otherwise "NO".
3 20 11 22 20 11 00 0
YESNO
思路:强联通分量模板题,如果强联通分量的个数等于点的个数,说明这个图根本不是个强联通图,恰好符合题意合法,输出YES,否则只要存在强联通分量则不合法
code:
#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;vector<int>G[102];vector<int>VG[102];vector<int>v;int vis[102];int n,m;void init(){int i;for(i = 0; i < n; i++){G[i].clear();VG[i].clear();}v.clear();memset(vis,0,sizeof(vis));}void dfs(int u){vis[u] = 1;int i;for(i = 0; i < G[u].size(); i++){if(!vis[G[u][i]]){dfs(G[u][i]);}}v.push_back(u);}void redfs(int u){vis[u] = 1;int i;for(i = 0; i < VG[u].size(); i++){if(!vis[VG[u][i]]){redfs(VG[u][i]);}}}int main(){while(~scanf("%d%d",&n,&m)){if(!n&&!m)break;init();int i;for(i = 0; i < m; i++){int a,b;cin >> a >> b; G[a].push_back(b);VG[b].push_back(a); }for(i = 0; i < n; i++){if(!vis[i]){dfs(i);}}memset(vis,0,sizeof(vis));int cnt = 0;for(i = v.size()-1; i >= 0; i--){if(!vis[v[i]]){redfs(v[i]);cnt++;}}if(n == cnt) cout << "YES" << endl;else cout << "NO" << endl; }return 0;}
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