uva 10305 - Ordering Tasks

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 <= 100and 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 withn = 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

纯拓扑排序，我按着刘汝佳那个写的，哈哈学习了。

#include<stdio.h>#include<stdlib.h>#include<string.h>#define maxn 105int topo[maxn],c[maxn],G[maxn][maxn],n,t;bool dfs(int u){     c[u]=-1; //-1表示正在访问中      for(int v=1;v<=n;v++)if(G[u][v])//查看是否存在关系      {          if(c[v]<0)return false;          if(!c[v]&&!dfs(v))return false;       }      c[u]=1;topo[--t]=u;//处理      return true;}bool toposort(){     t=n;     memset(c,0,sizeof(c));//初始化,0代表还未被访问      for(int i=1;i<=n;i++)        if(!c[i]&&!dfs(i))return false;     return true; }int main(){     int m,u,v;     while(scanf("%d%d",&n,&m)==2&&(m||n)){         memset(G,0,sizeof(G));          for(int i=1;i<=m;i++){             scanf("%d%d",&u,&v);             G[u][v]=1;          }         if(toposort()){//是否存在这样的拓扑排序，主要看有无有向环             for(int i=0;i<n-1;i++)printf("%d ",topo[i]);            printf("%d\n",topo[n-1]);          }      }     return 0;}