The Bottom of a Graph

Description

We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges. Then G=(V,E) is called a directed graph.
Let n be a positive integer, and let p=(e1,...,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,...,vn+1). Then p is called a path from vertex v1 to vertex vn+1 in G and we say that vn+1 is reachable from v1, writing (v1→vn+1).
Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w. The bottom of a graph is the subset of all nodes that are sinks, i.e., bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}. You have to calculate the bottom of certain graphs.

Input

The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer numbers in the set V={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integer e and, thereafter, e pairs of vertex identifiers v1,w1,...,ve,we with the meaning that (vi,wi)∈E. There are no edges other than specified by these pairs. The last test case is followed by a zero.

Output

For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line. Sample Input

`3 31 3 2 3 3 12 11 20`

Sample Output

```1 32题意：求出强连通分量，判断每个分量是否有出边。按序输出没出边的分量里的点。代码：
#include <stdio.h>#include <stdlib.h>#define Min(a, b) a < b ? a : b#define maxn 5001struct T{    int v, next;}fn[maxn * maxn];int g[maxn], visit[maxn], low[maxn], dfn[maxn], stack[maxn], f[maxn], flag[maxn];int top, id;void set(int n){    for (int i = 1; i <= n; i++)    {        g[i] = -1, visit[i] = 0, f[i] = 0, flag[i] = 0;    }}int dfs(int u, int t){    int i;    visit[u] = 1;    stack[top++] = u;    dfn[u] = low[u] = t;    for (i = g[u]; i != -1; i = fn[i].next)    {        if (!visit[fn[i].v])        {            t = dfs(fn[i].v, t + 1);//返回遍历完u的子树的时间             low[u] = Min(low[u], low[fn[i].v]);//如果有孩子因为后向边被缩小了low值，则它也相应缩小low值，整个分量的low都取分量中low最小的值。        }        else        {            if (!f[fn[i].v])//如果f[v]还是0，表示u->v是后向边，v还没出过栈过            {                low[u] = Min(low[u], low[fn[i].v]);            }        }    }    if (dfn[u] == low[u])    {        id++;        do        {            f[stack[--top]] = id;         }while (top > 0 && stack[top] != u);    }    return t;}void tarjan(int n){    int t = 0;    id = 0;    int i;    for (i = 1; i <= n; i++)    {        if (!visit[i])        {            //如果i点能与其它点x构成一个分量，则一定能搜到x，否则，它自己便是一个分量，所以这里的搜索不会重复。            t = dfs(i, t + 1);        }    }}int main(){    int n, m, u, v, th, k, i, j;    while (scanf("%d", &n), n)    {        scanf("%d", &m);        th =  0;        set(n);        while (m--)        {            scanf("%d%d", &u, &v);            fn[th].v = v, fn[th].next = g[u], g[u] = th++;        }        tarjan(n);        for (i = 1; i <= n; i++)        {             for (j = g[i]; j != -1; j = fn[j].next)            {                if (f[fn[j].v] != f[i])//两个点的所在分量不同，表示i有出边                 {                    //hash[i] = 1;                    flag[f[i]] = 1;                    break;                }            }        }        for (i = 1; i <= n; i++)        {            if (!flag[f[i]])            {                if (i != n)                {                    printf("%d ", i);                }                else                {                    printf("%d", i);                }            }        }        printf("\n");    }    system("pause");    return 0;}/*6 81 31 23 55 62 44 63 44 14 51 22 33 11 33 4*/
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