UVa 10305 Ordering Tasks(拓扑排序)

题意  输出n个数m组小于关系的一种可能的拓扑排序

应用dfs拓扑排序  访问j时  若存在i<j且i没被访问  就访问i

#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int n, m, t, v[N], tpo[N], g[N][N];

void dfs(int j)
{
    if(v[j]) return;
    for(int i = 1; i <= n; ++i)
        if(g[i][j]) dfs(i);
    v[j] = 1;
    tpo[++t] = j;
}

int main()
{
    int a, b, i;
    while(~scanf("%d %d", &n, &m), n)
    {
        memset(g, 0, sizeof(g));
        memset(v, 0, sizeof(v));
        while(m--)
        {
            scanf("%d%d", &a, &b);
            g[a][b] = 1;
        }

        t = 0;
        for(i = 1; i <= n; ++i)
            if(!v[i]) dfs(i);
        for(i = 1; i < n; ++i)
            printf("%d ", tpo[i]);
        printf("%d\n", tpo[n]);
    }
    return 0;
}

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 = 0will 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

UVa 10305 Ordering Tasks(拓扑排序)

原文：http://blog.csdn.net/acvay/article/details/43056457