M

Description

John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task is only possible if other tasks have already been executed.

Input

The input will consist of several instances of the problem. Each instance begins with a line containing two integers, 1 <= n <= 100 and m. n is the number of tasks (numbered from 1 to n) and m is the number of direct precedence relations between tasks. After this, there will be m lines with two integers i and j, representing the fact that task i must be executed before task j. An instance with n = m = 0 will finish the input.

Output

For each instance, print a line with n integers representing the tasks in a possible order of execution.

Sample Input

	5 4

	1 2

	2 3

	1 3

	1 5

	0 0

Sample Output

	1 4 2 5 3


解题思路：拓扑排序加上dfs。不错的一道题

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int ss[101][101],num[101],daan[101],m,n,t;
bool bfs(int i)
{
  num[i]=-1;
  for(int t=1;t<=m;t++)
          if(ss[i][t]==1)
          {
               if(num[t]==-1)return false;
               else if(!num[t]) bfs(t);
          }
          num[i]=1;daan[t--]=i;
          return true;
}
bool bbb()
{
  t=m;
  memset(num,0,sizeof(num));
  for(int y=1;y<=m;y++)
    if(!num[y]&&!bfs(y))
     return false;
     return true;
}
int main()
{
     int a,b;
     while(scanf("%d%d",&m,&n)==2)
     {    memset(ss,0,sizeof(ss));
          if(m==0&&n==0)break;
          for(int i=1;i<=n;i++)
          {
               scanf("%d%d",&a,&b);
               ss[a][b]=1;
          }
          if(bbb())
          {
               for(int z=1;z<m;z++)
                printf("%d ",daan[z]);
                 printf("%d
",daan[m]);
          }
             else printf("No
");
     }
     return 0;
}