Ordering Tasks_ordering tasks john has n tasks to do. unfortunate-优快云博客

本文链接：https://blog.youkuaiyun.com/Rechard_chen/article/details/41985651

本文介绍了一个基于深度优先搜索实现的拓扑排序算法。输入包括任务数量及任务间的依赖关系，输出为符合依赖关系的任务执行顺序。若存在环，则无法完成所有任务。

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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

题意：按拓扑排序输出;

<span style="font-size:14px;"># include <cstdio>
# include <cstring>
# include <iostream>
using namespace std;

int dp[1100][1100],vis[1100],ans[1100],n,m,t;

bool dfs(int i)
{
    vis[i]=-1;
    for(int j=1;j<=n;j++)    if(dp[i][j])    //判断两者是否联通
    {
        if(vis[j]<0)    return false;        //判断是否出现环
        else  if(!vis[j])    dfs(j);         //继续深搜
    }
    vis[i]=1;   ans[t--]=i;                  //标记
    return true;
}

bool toposort()     //以每一个点开始深搜；
{
    t=n;
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
        if(!vis[i])    if(!dfs(i))    return false;
    return true;
}

int main()
{
    //freopen("a.txt","r",stdin);
    while(cin>>n>>m&&n)
    {
        int i,x,y;
        memset(dp,0,sizeof(dp));
        for(i=0;i<m;i++)
        {
            scanf("%d%d",&x,&y);
            dp[x][y]=1;
        }
        if(toposort())
        {
            for(i=1;i<n;i++)
                printf("%d ", ans[i]);
            printf("%d\n", ans[n]);
        }
        else    printf("No\n");
    }
    return 0;
}</span>