（拓扑排序） uva 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.

题目大意：

约翰有n个任务要做， 不幸的是，这些任务并不是独立的，执行某个任务之前要先执行完其他相关联的任务。

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

拓扑排序的定义：

          对一个有向无环图(Directed Acyclic Graph简称DAG)G进行拓扑排序，是将G中所有顶点排成一个线性序列，使得图中任意一对顶点u和v，若<u，v> ∈E(G)，则u在线性序列中出现在v之前。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int G[110][110],n,m,a,b;
int vis[110],c[110], topo[110], t;

bool dfs(int u){
    vis[u] = -1; //表示正在访问
    for(int v=1; v<=n; ++v) if(G[u][v]){
        if(vis[v] == -1) return false; // 如果存在有向环，失败退出
        else if(!vis[v] && !dfs(v)) return false;
    }
    // 结束访问
    vis[u] = 1; topo[--t] = u;
    return true;
}

bool topoSort(){
    t = n;
    memset(vis, 0, sizeof(vis));
    for(int u=1; u<=n; ++u)
        if(!vis[u] && !dfs(u)) return false;
    return true;
}

int main(){

    while(~scanf("%d %d",&n,&m) && n+m){
        memset(G, 0, sizeof(G));
        for(int i=0; i<m; ++i){
            scanf("%d %d",&a,&b);
            G[a][b] = 1;
        }
        if(topoSort()) {
            printf("%d",topo[0]);
            for(int i=1; i<n; ++i)
                printf(" %d",topo[i]);
            printf("\n");
        }
    }
    return 0;
}