poj 2762 Going from u to v or from v to u 判断图中任意两点x,y是否总是存在x->y或者y->x

本文探讨了如何通过Kosaraju算法解决有向图中的强连通分量问题，并通过缩点和拓扑排序判断是否满足特定条件。适用于理解复杂有向图的性质和应用。

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
/*
给定一个有向图，求：

1) 至少要选几个顶点，才能做到从这些顶点出发，可以到达全部顶点
DAG上入度为0的点个数
2) 至少要加多少条边，才能使得从任何一个顶点出发，都能到达全部顶点
DAG上max(入度为0的点数,出度为0的点数）；
n=1时单独考虑
*/
const int maxn=100000;
struct edge
{
int t,w;//u->t=w;
int next;
};
int V,E;//点数(从1开始)，边数
int p[maxn],pf[maxn];//邻接表原图,逆图
edge G[maxn],Gf[maxn];//邻接表原图,逆图
int l,lf;
void init()
{
memset(p,-1,sizeof(p));
memset(pf,-1,sizeof(pf));
l=lf=0;
}
void addedge(int u,int t,int w,int l)
{
G[l].w=w;
G[l].t=t;
G[l].next=p[u];
p[u]=l;
}
void addedgef(int u,int t,int w,int l)
{
Gf[l].w=w;
Gf[l].t=t;
Gf[l].next=pf[u];
pf[u]=l;
}

///Kosaraju算法,返回为强连通分量个数
bool flag[maxn]; //访问标志数组
int belg[maxn]; //存储强连通分量,其中belg[i]表示顶点i属于第belg[i]个强连通分量
int numb[maxn]; //结束时间(出栈顺序)标记,其中numb[i]表示离开时间为i的顶点
//用于第一次深搜,求得numb[1..n]的值
void VisitOne(int cur, int &sig)
{
flag[cur] = true;
for (int i=p[cur];i!=-1;i=G[i].next)
{
if (!flag[G[i].t])
{
VisitOne(G[i].t,sig);
}
}
numb[++sig] = cur;
}
//用于第二次深搜,求得belg[1..n]的值
void VisitTwo(int cur, int sig)
{
flag[cur] = true;
belg[cur] = sig;
for (int i=pf[cur];i!=-1;i=Gf[i].next)
{
if (!flag[Gf[i].t])
{
VisitTwo(Gf[i].t,sig);
}
}
}
//Kosaraju算法,返回为强连通分量个数
int Kosaraju_StronglyConnectedComponent()
{
int i, sig;
//第一次深搜
memset(flag,0,sizeof(flag));
for ( sig=0,i=1; i<=V; ++i )
{
if ( false==flag[i] )
{
VisitOne(i,sig);
}
}
//第二次深搜
memset(flag,0,sizeof(flag));
for ( sig=0,i=V; i>0; --i )
{
if ( false==flag[numb[i]] )
{
VisitTwo(numb[i],++sig);
}
}
return sig;
}
//缩点
int n;//缩点后的点个数1~n
int g[maxn];
edge eg[maxn];//邻接表
int re;
int in[maxn];//入度
int mat[1100][1100];//DAG的邻接矩阵
void dinit(int sig)
{
memset(g,-1,sizeof(g));
n=sig;
re=0;
memset(mat,0,sizeof(mat));
memset(in,0,sizeof(in));
}
void addedge0(int u,int t,int w,int l)
{
eg[l].w=w;
eg[l].t=t;
eg[l].next=g[u];
g[u]=l;
}
int topo[maxn];
int ct;
bool toposort()//缩点后拓扑排序肯定不存在环
{
ct=0;
int top=-1;
for(int i=1;i<=n;i++)
{
if(in[i]==0) in[i]=top,top=i;
}
for(int l=1;l<=n;l++)
{
int cur=top;
//if(top==-1) return 0;
top=in[top];
topo[ct++]=cur;
for(int i=g[cur];i!=-1;i=eg[i].next)
{
in[eg[i].t]--;
if(in[eg[i].t]==0) in[eg[i].t]=top,top=eg[i].t;
}
}
return 1;
}
int main()
{
int ci;scanf("%d",&ci);
while(ci--)
{
init();
scanf("%d%d",&V,&E);
for(int i=1;i<=E;i++)
{
int u,t,w=1;scanf("%d%d",&u,&t);
addedge(u,t,w,l++);
addedgef(t,u,w,lf++);
}
int ans=Kosaraju_StronglyConnectedComponent();
//printf("%d/n",ans);
///缩点
dinit(ans);
for(int i=1;i<=V;i++)//构图
{
for(int j=p[i];j!=-1;j=G[j].next)
{
int st=belg[i],ed=belg[G[j].t];
if(st!=ed)
{
int flag=1;
for(int k=g[st];k!=-1;k=eg[k].next)
{
if(eg[k].t==ed)
{
flag=0;
break;
}
}
if(flag)
{
addedge0(st,ed,1,re++);
mat[st][ed]=1;
in[ed]++;
}
}
}
}
//拓扑排序如果拓扑序中任意第i个和第i+1个有边相连则YES 否则NO
toposort();
int flag=1;
for(int i=0;i<ct-1;i++)
{
int u=topo[i],v=topo[i+1];
if(!mat[u][v])
{
flag=0;
break;
}
}
if(flag) printf("Yes/n");
else printf("No/n");
}
return 0;
}

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

Sample Output

Yes

Description

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

Sample Output

Yes

Description

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