HDU 3342 Legal or Not（判断环）

Problem Description

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

3 2

0 1

1 2

2 2

0 1

1 0

0 0

Sample Output

YES

NO

判断是否有环

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) ((a,0,sizeof(a)))
typedef long long ll;
vector<int>v[];
int x,y,degree[];
int vis[],n,m;
void toposort()
{
    set<int>s;
    priority_queue<int,vector<int>,greater<int> >q;
    for(int i=;i<n;i++)
    {
        if(!degree[i])
        {
            q.push(i);
            vis[i]=;
        }
    }
    bool flag=;
    while(!q.empty())
    {
        int u=q.top();
        q.pop();
        s.insert(u);
        for(int i=;i<v[u].size();i++)
        {
            degree[v[u][i]]--;
            if(!degree[v[u][i]]) q.push(v[u][i]);
        }
    }
    puts(s.size()==n?"YES":"NO");
}
int main()
{
    while(scanf("%d%d",&n,&m)&& n+m)
    {
        for(int i=;i<=n;i++)
            v[i].clear();
        memset(degree,,sizeof(degree));
        for(int i=;i<m;i++)
        {
            scanf("%d%d",&x,&y);
            v[x].push_back(y);
            degree[y]++;
        }
        toposort();
    }
    return ;
}