Legal or Not HDU - 3342（强连通分量）

最新推荐文章于 2020-07-15 16:57:11 发布

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本文介绍了一种利用强联通分量算法判断图中“师傅”与“徒弟”关系是否合法的方法。通过构建图模型，使用深度优先搜索算法进行强联通分量的计算，若强联通分量的数量等于节点数量，则表明图中不存在非法的互为“师傅”与“徒弟”的情况。

Legal or Not

HDU - 3342

ACM-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.

Input

The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y's master and y is x's prentice. The input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.

Output

For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".

Sample Input

Sample Output

YES
NO

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;
}