By yusijia

Updated:

Contents 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63

#include <cstdio>
#include <cstring>
#define MAXN 100005
#define MAXM 100005

int edge[MAXM], head[MAXN], next[MAXM], edsz, n, m, indo[MAXN], ans[MAXN];

bool vis[MAXN];

inline void init()
{
    memset(vis, false, sizeof(vis));
    memset(indo, 0, sizeof(indo));
    memset(head, -1, sizeof(head));
    edsz = 0;
}

void adds(int a, int b)
{
    ++indo[b];
    edge[edsz] = b;
    next[edsz] = head[a];
    head[a] = edsz++;
}

void dfs(int k)
{
    if (k == n)
    {
        for (int i = 0; i < n - 1; ++i)
            printf("%d ", ans[i]);
        printf("%d\n", ans[n - 1]);
        return;
    }
    for (int i = 1; i <= n; ++i) if (!vis[i] && !indo[i])//这里类似全排列
    {//如果入度为0，且没访问过
        ans[k] = i;//跑遍第k个位置的所有可能
        vis[i] = true;
        for (int j = head[i]; ~j; j = next[j])
            --indo[edge[j]];
        dfs(k + 1);
        for (int j = head[i]; ~j; j = next[j])//恢复原来的状态
            ++indo[edge[j]];
        vis[i] = false;
    }
}

int main(int argc, char* argv[])
{
    int a, b;
    while (scanf("%d%d", &n, &m) != EOF)
    {
        init();
        for (int i = 0; i < m; ++i)
        {
            scanf("%d%d", &a, &b);
            adds(a, b);
        }
        dfs(0);
    }
    return 0;
}
Contents