Cactus

 

Problem Description

1. It is a Strongly Connected graph.
2. Each edge of the graph belongs to a circle and only belongs to one circle.
We call this graph as CACTUS.



There is an example as the figure above. The left one is a cactus, but the right one isn’t. Because the edge (0, 1) in the right graph belongs to two circles as (0, 1, 3) and (0, 1, 2, 3).

 

 

Input

The input consists of several test cases. The first line contains an integer T (1<=T<=10), representing the number of test cases.
For each case, the first line contains a integer n (1<=n<=20000), representing the number of points.
The following lines, each line has two numbers a and b, representing a single-way edge (a->b). Each case ends with (0 0).
Notice: The total number of edges does not exceed 50000.

 

 

Output

For each case, output a line contains “YES” or “NO”, representing whether this graph is a cactus or not.

Sample Input

2

4

0 1

1 2

2 0

2 3

3 2

0 0

4

0 1

1 2

2 3

3 0

1 3

0 0

 

Sample Output

YES

NO

 

 

 

 

 

 

 

 

 

 

一个图是仙人掌图必须满足条件：必须是强连通图，由一个或多个圆圈粘结而成。

算法：遍历图中每条边，发现圆圈立即标记圆圈上各点，检查是否有多个圆圈共边。

#include<iostream>
//#include<fstream> 
using namespace std;
//ifstream fin("fin.in");
//ofstream fout("fout.out"); 
 
typedef struct{int u,v,next;}Node;

Node edg[50001];int next[20001],tot=0; //tot edgs 

int t,n; 

void Update(int u,int v){
     edg[tot].u=u;edg[tot].v=v;edg[tot].next=next[u];next[u]=tot++; 
     } 
           
void Init(){
     memset(next,-1,sizeof(next));
     cin>>n; tot=0; 
     int a,b; 
     while(cin>>a>>b,a||b)
     Update(a,b);  
     } 
 
int stack[20020],queue[20020],st,qu,start;bool vis[20020],mark[20020];
 
int Check(){
    int u,v;
    memset(vis,0,sizeof(vis));
    memset(mark,0,sizeof(mark));
     
    queue[0]=0;st=0;qu=1; 
     
    while(qu>0)
    {
       --qu;start=u=queue[qu]; 
       if(next[u]==-1) continue;

       stack[st++]=u;vis[u]=1; 
                      
       while(next[u]!=-1) 
       {
        v=edg[next[u]].v; next[u]=edg[next[u]].next;
        if(v!=start&&mark[v]) return 0;
        
        if(!vis[v])
        {u=v;vis[u]=1;queue[++qu]=u;stack[st++]=u;vis[u]=1;}       
        else 
        {
         while(stack[st-1]!=v)
         {mark[stack[st-1]]=1;st--;} 
         u=v;    
                         }                 
                                   } 
                                           
                   
       if(next[u]==-1)
       {
        if(u==start)
        {st--;mark[u]=1;} 
        else return 0; 
                               }                         
                                           

                 } 
     return 1; 
     } 
 
 int main()
 {
     cin>>t;
     while(t>0)
     {
      t--; 
      Init(); 
      if(Check()) cout<<"YES"<<endl;
      else cout<<"NO"<<endl; 
               } 
     }