Ordering Tasks - UVa 10305 拓扑排序-优快云博客

本文介绍了一个典型的任务排序问题，即根据任务间的依赖关系确定合理的执行顺序。通过使用拓扑排序算法解决此类问题，并提供了完整的实现代码。

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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 andm. 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

题意：一个任务需要已经做完某些任务才能进行，设计任务的排序。

思路：简单的拓扑排序裸题。每次找到入度为0的点，同时删除以它为出度的边。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int n,m,G[110][110],vis[110];
bool check(int u)
{
    int i;
    for(i=1;i<=n;i++)
       if(G[i][u])
         return false;
    return true;
}
void solve(int u)
{
    int i;
    for(i=1;i<=n;i++)
       G[u][i]=0;
}
int main()
{
    int i,j,k,ret,u,v;
    while(~scanf("%d %d",&n,&m) && n+m)
    {
        memset(G,0,sizeof(G));
        memset(vis,0,sizeof(vis));
        ret=0;
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&u,&v);
            G[u][v]=1;
        }
        while(ret<n)
        {
            for(i=1;i<=n;i++)
               if(!vis[i])
               {
                   if(check(i))
                   {
                       vis[i]=1;
                       if(ret)
                         printf(" ");
                       printf("%d",i);
                       ret++;
                       solve(i);
                   }
               }
        }
        printf("\n");
    }
}