HDOJ 题目3594 Cactus（强连通，判仙人掌图）

最新推荐文章于 2018-11-27 20:45:48 发布

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本文介绍了一种用于判断给定图是否为仙人掌图的算法，并提供了详细的实现代码。仙人掌图是一种强连通图，其特点在于每条边恰好属于一个环。文章通过示例阐述了仙人掌图的概念，并给出了输入输出样例。

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Cactus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1365 Accepted Submission(s): 654

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

Sample Output

  
   YES
NO

Author

alpc91

Source

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

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ac代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
#include<string>
#define INF 0xfffffff
#define min(a,b) (a>b?b:a)
using namespace std;
int dfn[20005*3],low[20005*3],stack[20005*3],head[20005*3],ins[20005*3],belong[20005*3];
int cnt,top,time,n,taj,flag;
struct s
{
	int u,v,next;
}edge[50005*3];
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void init()
{
	memset(dfn,-1,sizeof(dfn));
	memset(stack,0,sizeof(stack));
	memset(low,-1,sizeof(low));
	memset(head,-1,sizeof(head));
	memset(ins,0,sizeof(ins));
	memset(belong,-1,sizeof(belong));
	cnt=top=time=taj=0;
}
void tarjan(int u)
{
	dfn[u]=low[u]=time++;
	ins[u]=1;
	stack[top++]=u;
	for(int i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].v;
		if(dfn[v]==-1)
		{
			tarjan(v);
			low[u]=min(low[v],low[u]);
		}
		else
		{	
			if(ins[v])
			{
				low[u]=min(dfn[v],low[u]);
				if(dfn[v]!=low[v])
				{
					flag=1;
				}
			}
		}
	}
	if(low[u]==dfn[u])
	{
		int now;
		taj++;
		do
		{
			now=stack[--top];
			ins[now]=0;
			if(belong[now]!=-1)
				flag=1;
			belong[now]=taj;
		}while(now!=u);
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int a,b;
		scanf("%d",&n);
		init();
		while(scanf("%d%d",&a,&b),a||b)
		{
			add(a,b);
		}
		flag=0;
		for(int i=0;i<n;i++)
		{
			if(dfn[i]==-1)
				tarjan(i);
		}
		if(taj==1&&flag==0)
			printf("YES\n");
		else
			printf("NO\n");
	}
}