POJ 2762 Going from u to v or from v to u?（强连通+拓扑）

POJ 2762 Going from u to v or from v to u?

题目链接

题意：给定一些有向图，要判断该图是否满足任意两点，要么u能到v，要么v能到u

思路：先缩点，然后拓扑排序，排序过程中如果队列中有任意时刻点大于2个，就代表出现分支了，肯定就不行了

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <stack>
#include <queue>
using namespace std;

const int N = 1005;
int t, n, m, vis[N][N];
vector<int> g[N], scc[N];
stack<int> S;

int pre[N], dfn[N], dfs_clock, sccn, sccno[N];

void dfs_scc(int u) {
	pre[u] = dfn[u] = ++dfs_clock;
	S.push(u);
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (!pre[v]) {
			dfs_scc(v);
			dfn[u] = min(dfn[u], dfn[v]);
		} else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);
	}
	if (dfn[u] == pre[u]) {
		sccn++;
		while (1) {
			int x = S.top(); S.pop();
			sccno[x] = sccn;
			if (x == u) break;
		}
	}
}

void find_scc() {
	sccn = dfs_clock = 0;
	memset(pre, 0, sizeof(pre));
	memset(sccno, 0, sizeof(sccno));
	for (int i = 1; i <= n; i++)
		if (!pre[i]) dfs_scc(i);
}

int in[N];

int main() {
	scanf("%d", &t);
	while (t) {
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) g[i].clear();
		int u, v;
		while (m--) {
			scanf("%d%d", &u, &v);
			g[u].push_back(v);
		}
		find_scc();
		memset(in, 0, sizeof(in));
		for (int i = 1; i <= sccn; i++) scc[i].clear();
		for (int u = 1; u <= n; u++) {
			for (int j = 0; j < g[u].size(); j++) {
				int v = g[u][j];
				int su = sccno[u], sv = sccno[v];
				if (su == sv || vis[su][sv] == t) continue;
				vis[su][sv] = t;
				in[sv]++;
				scc[su].push_back(sv);
			}
		}
		queue<int> Q;
		for (int i = 1; i <= sccn; i++)
			if (!in[i]) Q.push(i);
		while (Q.size() == 1) {
			int u = Q.front();
			Q.pop();
			for (int i = 0; i < scc[u].size(); i++) {
				int v = scc[u][i];
				in[v]--;
				if (in[v] == 0) Q.push(v);
			}
		}
		printf("%s\n", Q.empty() ? "Yes" : "No");
		t--;
	}
	return 0;
}