BOTTOM - The Bottom of a Graph

考虑缩点。

发现对于每一个强连通分量的每个点，都满足能到其所在强连通分量的每个点然后走回来。

但是这个点能到其他强连通分量的点走回来吗？并不是的。由于缩点完是有向无环图，所以只有缩点后出度为 $0$ 的强连通分量内的所有点满足题意。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 50005;
vector<int> G[N];
int dfn[N], low[N], stk[N], top, idx;
bool in_stk[N];
int in[N], scc_cnt, id[N];
vector<int> ve[N];
int n, m;
int u[N], v[N];

void tarjan(int u)
{
	::dfn[u] = ::low[u] = ++::idx;
	::in_stk[u] = 1;
	::stk[++::top] = u;
	for (auto& j : ::G[u])
	{
		if (!::dfn[j])
		{
			tarjan(j);
			::low[u] = ::min(::low[u], ::low[j]);
		}
		else if (::in_stk[j]) ::low[u] = ::min(::low[u], ::dfn[j]);
	}
	if (::dfn[u] == ::low[u])
	{
		::scc_cnt++;
		int y = 0;
		do
		{
			y = ::stk[::top--];
			::in_stk[y] = 0;
			::id[y] = ::scc_cnt;
			::ve[::scc_cnt].emplace_back(y);
		} while (y != u);
	}
};

int main()
{
	while (scanf("%d", &n))
	{
		if (n == 0) break;
		::idx = 0;
		::top = ::scc_cnt = 0;
		scanf("%d", &m);
		for (int i = 1; i <= n; i++) ::G[i].clear(), ::in[i] = 0, ::in_stk[i] = 0, ::ve[i].clear(), ::dfn[i] = ::low[i] = 0;
		for (int i = 1; i <= m; i++)
		{
			int u, v;
			scanf("%d%d", &u, &v);
			G[u].emplace_back(v);
			::u[i] = u, ::v[i] = v;
		}
		for (int i = 1; i <= ::n; i++)
		{
			if (!::dfn[i]) ::tarjan(i);
		}
		for (int i = 1; i <= ::m; i++)
		{
			int u = ::u[i], v = ::v[i];
			if (::id[u] ^ ::id[v])
			{
				::in[::id[u]]++;
			}
		}
		vector<int> ans;
		for (int i = 1; i <= ::scc_cnt; i++)
		{
			if (::in[i] == 0)
			{
				for (auto& j : ::ve[i])
				{
					ans.emplace_back(j);
				}
			}
		}
		sort(ans.begin(), ans.end());
		for (auto& i : ans)
		{
			printf("%d ", i);
		}
		printf("\n");
	}
	return 0;
}