Cactus

Cactus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 56 Accepted Submission(s): 34

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

 

Author

alpc91

 

Source

2010 ACM-ICPC Multi-University Training Contest（16）——Host by NUDT

 

Recommend

zhengfeng

 

/*
题意：给你一个强连通图，如果每条边都只属于一个圈，那么这个图就是仙人掌图。让你判断一下是不是仙人掌图。

初步思路：tarjan遍历边，每次遇到以前的边的时候，这个环中每个边都标记一下如果这条边标记两次，那么就不是仙人掌图

*/
#include<bits/stdc++.h>
#define N 20005
#define M 50005
using namespace std;
/**************************强连通模板******************************/
struct node{
    int from,to,next;
}edge[M];
int tol,head[N],dfn[N],low[N],flag,Count,cnt,n;
bool visit[N],vis[N];
stack<int>S;
void add(int a,int b)
{
    edge[tol].from=a;edge[tol].to=b;edge[tol].next=head[a];head[a]=tol++;
}
int min(int a,int b)
{
    return a<b?a:b;
}
void tarjan(int u)
{
    int j,v;
    dfn[u]=low[u]=cnt++;
    visit[u]=vis[u]=1;
    S.push(u);
    for(j=head[u];j!=-1;j=edge[j].next)
    {
        v=edge[j].to;
        if(!visit[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(vis[v]) 
        {
            low[u]=min(low[u],dfn[v]);
            if(dfn[v]!=low[v]) flag=1; 
        }
        if(flag) return;
    }
    if(dfn[u]==low[u])
    {
        Count++;
        do{
            v=S.top();
            S.pop();
            vis[v]=0;
        }while(v!=u);
    }
}
/**************************强连通模板******************************/
int main()
{
    int i,ncase,a,b;
    scanf("%d",&ncase);
    while(ncase--)
    {
        scanf("%d",&n);
        tol=0;
        memset(head,-1,sizeof(head));
        while(scanf("%d%d",&a,&b)!=EOF)
        {
            if(!a && !b) break;
            add(a,b);
        }
        memset(visit,0,sizeof(visit));
        memset(vis,0,sizeof(vis));
        flag=0;
        Count=0;
        cnt=0;
        for(i=0;i<n;i++)
            if(!visit[i]) tarjan(i);
        if(flag||Count!=1) printf("NO
");
        else printf("YES
");
    }
    return 0;
}