HDOJ 5631 Rikka with Graph(并查集)

最新推荐文章于 2019-08-16 15:37:33 发布

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本文探讨了一个特定的图论问题：在给定一个包含n个顶点和n+1条边的非定向图的情况下，通过删除至少一条边来确保剩余的图仍保持连通性的不同方案的数量。文章提供了详细的解题思路及实现代码。

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Rikka with Graph

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

Total Submission(s): 172 Accepted Submission(s): 74

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has a non-direct graph with n vertices and n+1 edges. Rikka can choose some of the edges (at least one) and delete them from the graph.

Yuta wants to know the number of the ways to choose the edges in order to make the remaining graph connected.

It is too difficult for Rikka. Can you help her?

Input

The first line contains a number

T(T≤30)——The number of the testcases.

For each testcase, the first line contains a number

n(n≤100).

Then n+1 lines follow. Each line contains two numbers

u,v , which means there is an edge between u and v.

Output

For each testcase, print a single number.

Sample Input

Sample Output

题意：给一个n个点n+1条边的图，选择删去一些边（至少一条），有多少方案使得删边后任是连通图？

题解：有个n点的图，至少有n-1条边才能保证连通，所以我们枚举删去一条边和删去两条边的这两种情况的所有可能，对于每种可能我们用并查集判断是否连通。

代码如下：

#include<cstdio>
#include<cstring>
int u[110],v[110],n;
bool use[110];
int tree[110];

int find(int x)
{
	if(tree[x]==x)
		return x;
	else
		return tree[x]=find(tree[x]);
} 

void merge(int a,int b)
{
	int fa=find(a);
	int fb=find(b);
	if(fa!=fb)
		tree[fa]=fb;
}

bool judge()//判断是否连通 
{
	int i;
	for(i=1;i<=n;++i)
		tree[i]=i;
	for(i=0;i<=n;++i)
	{
		if(use[i])
			merge(u[i],v[i]);
	}
	int cnt=0;
	bool sign=true;
	for(i=1;i<=n;++i)
	{
		if(tree[i]==i)
			cnt++;
		if(cnt>1)
		{
			sign=false;
			break;
		}
	}
	return sign;
}

int main()
{
	int i,j,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=0;i<=n;++i)
		{
			scanf("%d%d",&u[i],&v[i]);
			use[i]=true;
		}
		int ans=0;
		for(i=0;i<=n;++i)
		{
			use[i]=false;
			if(judge())//删去一条边 
				ans++;
			for(j=i+1;j<=n;++j)
			{
				use[j]=false;
				if(judge())//删去两条边 
					ans++;
				use[j]=true;
			}
			use[i]=true;
		}
		printf("%d\n",ans);
	}
	return 0;
}