uva 10305 - Ordering Tasks

Problem F

Ordering Tasks

Input: standard input

Output: standard output

Time Limit:1 second

Memory Limit:32 MB

 

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 bem lines with two integers i andj, representing the fact that task i must be executed before taskj. 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

拓扑排序，每次找入度为0的点，删除相关边然后重复操作。

#include<stdio.h>int n,step,a[101][101],visit[101];int sort(){int i,x=0,y,f=1; if (step<=0) return 0; --step; while (f) {++x; y=0;  if (visit[x])  { for (i=1;i<=n;i++)    {y=y+a[i][x];     if (y>0) break;    }    f=y;  } } for (i=1;i<=n;i++) a[x][i]=0; visit[x]=0; printf("%d",x); if (step) printf(" ");      else printf("\n"); sort(); return 0;}int main(){int x,y,i,j,m; while (scanf("%d%d",&n,&m),n+m) {for (i=1;i<=n;i++)  for (j=1;j<=n;j++)  a[i][j]=0;  for (i=1;i<=n;i++)  visit[i]=1;  for (i=1;i<=m;i++)  {scanf("%d%d",&x,&y);   a[x][y]=1;  }  step=n;  sort(); } return 0;}