BC Harry and Magical Computer (拓扑排序)

Harry and Magical Computer

 Accepts: 350

 Submissions: 1348

 Time Limit: 2000/1000 MS (Java/Others)

 Memory Limit: 32768/32768 K (Java/Others)

Problem Description

In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

Input

There are several test cases, you should process to the end of file. For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. 1≤n≤100,1≤m≤10000 The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1≤a,b≤n

Output

Output one line for each test case. If the computer can finish all the process print "YES" (Without quotes). Else print "NO" (Without quotes).

Sample Input

3 2
3 1
2 1
3 3
3 2
2 1
1 3

Sample Output

YES
NO

拓扑排序模板题。注意在处理IN[i] ++的时候要先判断是否已经存在这条边。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <cctype>
 #include <cmath>
 #include <queue>
 #include <map>
 #include <cstdlib>
 using    namespace    std;

 const    int    SIZE = ;
 int    IN[SIZE];
 bool    G[SIZE][SIZE];
 int    N,M;

 bool    toposort(void);
 int    main(void)
 {
     int    from,to;

     while(scanf("%d%d",&N,&M) != EOF)
     {
         fill(IN,IN + SIZE,);
         fill(&G[][],&G[SIZE - ][SIZE - ],false);
         for(int i = ;i < M;i ++)
         {
             scanf("%d%d",&from,&to);
             if(!G[from][to])
                 IN[to] ++;
             G[from][to] = ;
         }
         if(toposort())
             puts("YES");
         else
             puts("NO");
     }

     return    ;
 }

 bool    toposort(void)
 {
     int    count = ;
     queue<int>    que;

     for(int i = ;i <= N;i ++)
         if(!IN[i])
             que.push(i);

     while(!que.empty())
     {
         int    cur = que.front();
         que.pop();
         count ++;

         for(int i = ;i <= N;i ++)
             if(G[cur][i])
             {
                 IN[i] --;
                 if(!IN[i])
                     que.push(i);
             }
     }
     if(count < N)
         return    false;
     return    true;
 }