題意：在v可以到達的所有點也都可以到達v,由此就可以知道求解縮點以後出度為0的點中的節點數字即可

思路：就是縮點後，輸出出度為0的點內的點。

#pragma GCC optimize(2)
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<set>
#include<vector>
#include<string>
#include<queue>
using namespace std;
const int maxn = 5005;
const int inf = 0x3f3f3f3f;
typedef long long ll;
int head[maxn], belong[maxn], Stack[maxn], instack[maxn], low[maxn], dfn[maxn];
int ou[maxn];
int n, m, tot, cnt;
int sum, top;
struct node
{
	int u, v, next;
}edge[maxn*maxn];
void addedge(int u, int v)
{
	edge[tot].u = u;
	edge[tot].v = v;
	edge[tot].next = head[u];
	head[u] = tot++;
	return;
}
void init()
{
	tot = cnt = 0;
	top = 0;
	sum = 0;
	memset(head, -1, sizeof(head));
	memset(dfn, 0, sizeof(dfn));
	memset(Stack, 0, sizeof(Stack));
	memset(instack, 0, sizeof(instack));
	memset(low, 0, sizeof(low));
	memset(belong, 0, sizeof(belong));
	memset(ou, 0, sizeof(ou));
	return;
}
void tarjan(int u)
{
	dfn[u] = low[u] = ++cnt;
	Stack[top++] = u;
	instack[u] = 1;
	for (int i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].v;
		if (!dfn[v])
		{
			tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else if (instack[v])
		{
			low[u] = min(low[u], dfn[v]);
		}
	}
	if (low[u] == dfn[u])
	{
		int v;
		sum++;
		do 
		{
			v = Stack[--top];
			instack[v] = 0;
			belong[v] = sum;
		} while (u!=v);
	}
}
int main()
{
	//freopen("C://input.txt", "r", stdin);
	while (scanf("%d%d", &n,&m)&&n!=0)
	{
		init();
		for (int i = 1; i <= m; i++)
		{
			int u, v;
			scanf("%d%d", &u, &v);
			addedge(u, v);
		}
		for (int i = 1; i <= n; i++)
		{
			if (!dfn[i])
			{
				tarjan(i);
			}
		}
		for (int i = 1; i <= n; i++)
		{
			for (int j = head[i]; j != -1; j = edge[j].next)
			{
				int v = edge[j].v;
				if (belong[i] != belong[v])
				{
					ou[belong[i]]++;
				}
			}
		}
		for (int i = 1; i <= n; i++)
		{
			if (!ou[belong[i]] )
			{
				printf("%d ", i);
			}
		}
		printf("\n");
	}
	return 0;
}