POJ 2762 Going from u to v or from v to u? / 强连通分量&&拓扑

给你一张图 判断是否任意2点u,v 满足要么u->v 可达 或者 v->u 可达 相互可达也可以

强连通分量缩点 在做拓扑 拓扑唯一 说明都互相可达

有空放弃用矩阵表示的拓扑 浪费时间 浪费空间

#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
const int maxn = 1010;

vector <int> G[maxn];
int pre[maxn];
int low[maxn];
int sccno[maxn];
int dfs_clock;
int scc_cnt;
stack <int> S;
int n, m;
int degree[maxn];
int cnt[maxn];
int Topo[maxn][maxn];
void dfs(int u)
{
	pre[u] = low[u] = ++dfs_clock;
	S.push(u);
	for(int i = 0; i < G[u].size(); i++)
	{
		int v = G[u][i];
		if(!pre[v])
		{
			dfs(v);
			low[u] = min(low[u], low[v]);
		}
		else if(!sccno[v])
			low[u] = min(low[u], pre[v]);
	}
	if(pre[u] == low[u])
	{
		scc_cnt++;
		while(1)
		{
			cnt[scc_cnt]++;
			int x = S.top();
			S.pop();
			sccno[x] = scc_cnt;
			if(x == u)
				break;
		}
	}
}
void find_scc()
{
	dfs_clock = scc_cnt = 0;
	memset(sccno, 0, sizeof(sccno));
	memset(pre, 0, sizeof(pre));
	memset(cnt, 0, sizeof(cnt));
	for(int i = 1; i <= n; i++)
		if(!pre[i])
			dfs(i);
}

bool topo()
{
	for(int i = 1; i <= scc_cnt; i++)
	{
		int ans = 0;
		for(int j = 1; j <= scc_cnt; j++)
		{
			if(!degree[j])
				ans++;
			if(ans > 1)
				return false;
		}
		for(int j = 1; j <= scc_cnt; j++)
		{
			if(!degree[j])
			{
				degree[j]--;	
				for(int k = 1; k <= scc_cnt; k++)
				{
					if(Topo[k][j])
						degree[k]--;
				}
				break;
			}
		}
	}
	return true;
}
int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d %d", &n, &m);
		for(int i = 1; i <= n; i++)
			G[i].clear();
		while(m--)
		{
			int u, v;
			scanf("%d %d", &u, &v);
			G[u].push_back(v);
		}
		find_scc();
		memset(degree, 0, sizeof(degree));
		memset(Topo, 0, sizeof(Topo));
		for(int i = 1; i <= n; i++)
		{
			for(int j = 0; j < G[i].size(); j++)
			{
				int v = G[i][j];
				if(sccno[i] != sccno[v])
				{
					degree[sccno[i]]++;
					Topo[sccno[i]][sccno[v]] = 1;
				}
			}
		}
		if(topo())
			printf("Yes\n");
		else
			printf("No\n");
	}
	return 0;
}