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

原题：http://poj.org/problem?id=2762

题意：给定n个点，m条有向边；

问是否满足图中任意两点单连通，即 u到v 或 v到u；



思路：首先用强连通缩点，然后直接判断是否为单链；

可以用拓扑的方式来判断，每次入度为0的点只能有一个；

#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<algorithm>

using namespace std;
const int N = 1100;
int head
;
bool vis
;
int DFN
, low
, stack
, belong
;
int n, m, top, taj, time, edgenum;

struct node
{
	int from, to, nex;
	bool sign;
}edge[6100];

void add(int u, int v)
{
	edge[edgenum].from = u;
	edge[edgenum].to = v;
	edge[edgenum].sign = false;
	edge[edgenum].nex = head[u];
	head[u] = edgenum++;
}

vector<int>bcc
;
void tarjan(int u, int fa)
{
	DFN[u] = low[u] = time++;
	vis[u] = 1;
	stack[top++] = u;
	for(int i = head[u];i!=-1;i = edge[i].nex)
	{
		int v = edge[i].to;
		if(DFN[v] == -1)
		{
			tarjan(v, u);
			low[u] = min(low[u], low[v]);
			if(DFN[u]<low[v])
				edge[i].sign = 1;
		}
		else if(vis[v])
		low[u] = min(low[u], DFN[v]);
	}
	if(low[u] == DFN[u])
	{  
		taj ++ ; 
		bcc[taj].clear();
		while(1)
		{
			int now = stack[--top];  
			vis[now] = 0; 
			belong[now] = taj;
			bcc[taj].push_back(now);
			if(now == u)
			break;
		}
	}
}

vector<int>G
;
int du
;
void suodian()
{
	memset(du, 0, sizeof(du));
	for(int i = 1;i<=taj;i++)
	G[i].clear();
	for(int i = 0;i<edgenum;i++)
	{
		int u = belong[edge[i].from];
		int v = belong[edge[i].to];
		if(u!=v)
		{
			G[u].push_back(v);
			du[v]++;
		}
	}
}

void tarjan_init(int all)
{
	memset(DFN, -1, sizeof(DFN));
	memset(low, -1, sizeof(low));
	memset(vis, 0, sizeof(vis));
	time = top = taj = 0;
	for(int i = 1;i<=all;i++)
	{
		if(DFN[i] == -1)
		tarjan(i, i);
	}
}

int main()
{
	int cas;
	scanf("%d", &cas);
	while(cas--)
	{
		scanf("%d%d", &n, &m);
		memset(head, -1, sizeof(head));
		edgenum = 0;
		while(m--)
		{
			int u, v;
			scanf("%d%d", &u, &v);
			add(u, v);
		}
		tarjan_init(n);
		suodian();
		int flag = 0;
		int ans
;
		memset(ans, 0, sizeof(ans));
		for(int t = 0;t<taj;t++)
		{
			int con = 0;
			for(int i = 1;i<=taj;i++)
			{
				if(du[i] == 0)
				{
					du[i] = -1;
					con++;
					for(int j = 0;j<G[i].size();j++)
					du[G[i][j]]--;
				}
			}
			if(con>1)
			{
				flag = 1;
				break;
			}
		}
		if(flag)
		printf("No\n");
		else
		printf("Yes\n");
	}
	return 0;
}