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

书上167的例题。。。不过我貌似没有按照上面的敲（然后书上考虑的比较周全）

#include

using namespace std;

const int maxn = 100 + 5;
int n, m, tmp;
int atlas[maxn][maxn];
int ind[maxn];
bool vis[maxn];

void dfs(int s){
    for (int i = 1; i <= n; i++){
        if (atlas[s][i]){
            if (!vis[i]){
                vis[i] = true;
                dfs(i);
                ind[--tmp] = i;
            }
        }
    }
}

int main(){
    while (scanf("%d%d", &n, &m) && n + m != 0){
        memset(atlas, 0, sizeof(atlas));
        for (int i = 0; i < m; i++){
            int u, v;
            scanf("%d%d", &u, &v);
            atlas[u][v] = 1;
        }
        tmp = n;
        memset(ind, -1, sizeof(ind));
        memset(vis, false ,sizeof(vis));
        for (int i = 1; i <= n; i++){
            if (vis[i] == true) continue;
            vis[i] = true;
            //if (!dfs(i)) continue;
            dfs(i);
            ind[--tmp] = i;
        }
        for (int i = tmp; i < n; i++){
            printf("%d%c", ind[i], i == n-1 ? '\n' : ' ');
        }
    }
    return 0;
}