Going from u to v or from v to u?_POJ2762强连通+并查集缩点+拓扑排序

     Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one
to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair
of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

The first line contains a single integer T, the number of test cases. And followed T cases.



The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly.

Output

The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes

Source

POJ Monthly--2006.02.26,zgl & twb





题意：jiajia有一个洞穴，洞穴中有n个房间，房间的连接是有方向的,jiajia想知道u与v之间是不是有路径u->v或v->u





思路：单连通问题，首先将图中的强连通的部分进行缩点，构成一颗树，接下来拓扑排序，判断拓扑排序的两点之间是不是有边，如果没有边则图不符合要求。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <algorithm>
 
using namespace std;
 
const int Max = 1100;
 
const int INF = 0x3f3f3f3f;
 
typedef struct node
{
	int v;
 
	int next;
 
}Line ; 
 
Line Li[Max*6];
 
int Head[Max],top;
 
int Map[Max][Max];
 
int Du[Max],pre[Max];
 
int vis1[Max],dfn[Max],low[Max];
 
bool vis2[Max];
 
int dep;
 
int Point[Max*5][2],Num;
 
int topo[Max],num;
 
int a[Max],ToNum;
 
stack < int >S;
 
void AddEdge(int u,int v)
{
	Li[top].v = v; Li[top].next = Head[u];
 
	Head[u] = top++;
}
 
int Find(int x)
{
	return pre[x]==-1?x:pre[x] = Find(pre[x]);
}
 
void Tarjan(int u) //强连通缩点
{
 
	dfn[u] = low[u] = dep++;
 
	vis1[u] = 1;
 
	S.push(u);
 
	for(int i=Head[u];i!=-1;i=Li[i].next)
	{
		if(vis1[Li[i].v]==1)
		{
			low[u]=min(low[u],dfn[Li[i].v]);
		}
		if(vis1[Li[i].v]==0)
		{
			Tarjan(Li[i].v);
 
			low[u]=min(low[u],low[Li[i].v]);
		}
 
		if(vis2[Li[i].v])
		{
			Point[Num][0]=u;
 
			Point[Num++][1]=Li[i].v;
		}
	}
 
	if(low[u]==dfn[u]) //如果low[u]==dfn[u],则说明是强连通的根节点。
	{
		vis2[u]=true;
 
		topo[num++] = u;
 
		while(1)
		{
			if(S.empty())
			{
				break;
			}
			int v = S.top();
 
			S.pop();
 
			vis1[v]=2;
 
			if(v==u)
			{
				break;
			}
			pre[v]=u;
 
		}
	}
}
 
void Toposort()//BFS拓扑排序
{
 
	queue<int>Q;
	for(int i=0;i<num;i++)
	{
		if(Du[topo[i]]==0)
		{
			Q.push(topo[i]);
 
		}
	}
	while(!Q.empty())
	{
		int u=Q.front();
 
		a[ToNum++]=u;
 
		Q.pop();
 
		for(int i=0;i<num;i++)
		{
			if(Map[u][topo[i]])
			{
				Du[topo[i]]--;
 
				if(Du[topo[i]]==0)
				{
					Q.push(topo[i]);
				}
			}
		}
	}
}
 
int main()
{
 
	int T;
 
	int n,m;
 
	scanf("%d",&T);
 
	while(T--)
	{
		scanf("%d %d",&n,&m);
 
		top =  0;
 
		memset(Head,-1,sizeof(Head));
 
		int u,v;
 
		for(int i=0;i<m;i++)
		{
			scanf("%d %d",&u,&v);
 
			AddEdge(u,v);
		}
 
		memset(vis1,0,sizeof(vis1));
 
		memset(Map,0,sizeof(Map));
 
		memset(vis2,false,sizeof(vis2));
 
		memset(Du,0,sizeof(Du));
 
		memset(pre,-1,sizeof(pre));
 
		dep  = 0; Num  =0 ;num = 0;
 
		while(!S.empty())
		{
			S.pop();
		}
 
		for(int i=1;i<=n;i++)
		{
			if(vis1[i]==0)
			{
				Tarjan(i);
			}
		}
		for(int i=0;i<Num;i++)
		{
 
			int x = Find(Point[i][0]);
 
			int y = Find(Point[i][1]);
 
			Map[x][y]=1;
 
			Du[y]++;
		}
 
		ToNum = 0;
 
		Toposort();
 
		bool flag=false;
 
		for(int i=0;i<ToNum-1;i++)
		{
			if(!Map[a[i]][a[i+1]])//判断相邻的是不是存在边
			{
				flag=true;
 
				break;
			}
		}
 
		if(flag)
		{
			printf("No\n");
		}
		else
		{
			printf("Yes\n");
		}
 
	}
 
	return 0;
}