題目描述

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.

輸入

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.

輸出

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

樣例輸入

4 3
0 1
1 2
2 3
3 3
0 1
1 2
2 0
0 1

樣例輸出

YES
NO

分析：給出 n 個點，m 條邊的圖，問這個圖是否是有向無環圖。進行拓撲排序即可。

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0x3fffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;bool TopologicalSort(int in[],int n,vector<int>dag[])
{
//    bool vis[n+5]={0};int cnt=0;queue<int>que;for(int i=0;i<n;++i){if(in[i]==0)que.push(i);}while(!que.empty()){int temp=que.front();que.pop();cnt++;int len=dag[temp].size();for(int i=0;i<len;++i){int index=dag[temp][i];in[index]--;if(in[index]==0)que.push(index);}}if(cnt==n)return true;return false;
}int main(void)
{#ifdef testfreopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);clock_t start=clock();#endif //testint n,m;while(scanf("%d%d",&n,&m),n){int in[n+5]={0};vector<int>dag[n+5];for(int i=0;i<m;++i){int a,b;scanf("%d%d",&a,&b);dag[a].push_back(b);in[b]++;}bool ans=TopologicalSort(in,n,dag);if(ans==1)printf("YES\n");else printf("NO\n");}#ifdef testclockid_t end=clock();double endtime=(double)(end-start)/CLOCKS_PER_SEC;printf("\n\n\n\n\n");cout<<"Total time:"<<endtime<<"s"<<endl;        //s為單位cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms為單位#endif //testreturn 0;
}