Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K

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?

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.

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

1

3 3

1 2

2 3

3 1

Yes

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;

}