【杭电oj5631】Rikka with Graph

最新推荐文章于 2024-11-19 20:38:50 发布

原创最新推荐文章于 2024-11-19 20:38:50 发布 · 314 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#并查集

HDOJ 同时被 2 个专栏收录

111 篇文章

订阅专栏

并查集

3 篇文章

订阅专栏

本文探讨了一个特定的图论问题：如何从含有n+1条边的非定向图中选择部分边删除，使得剩余的图保持连通状态。通过详细解释算法思路并给出实现代码，为读者提供了理解和解决此类问题的方法。

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 999 Accepted Submission(s): 468

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

  
   9
  
  

  
  

  
  

  
  
   要连接所有节点至少需要n-1条线，而总共的线数是n+1（题目给的），给的数据又比较小，所以暴力把所有去掉一条线和去掉两条线的情况遍历一次。求出能连通所有节点的情况的总和。
  
  

  
  

  
  

  
  

  
  
   #include<stdio.h>
const int N = 111;
int a[N][2];
int bin[N];
int find(int x) {

   	int r=x;

   	while(bin[r]!=r)

   		r=bin[r];

   	return r;
}
void Union(int x,int y) {

   	int nx=find(x);

   	int ny=find(y);

   	if(nx!=ny)

   		bin[nx]=ny;
}
int main() {

   	int T;

   	scanf("%d",&T);

   	while(T--) {

   		int n,ans=0,t;

   		scanf("%d",&n);

   		for(int l=1; l<=n+1; l++)

   			scanf("%d%d",&a[l][0],&a[l][1]);

   		for(int j=1; j<=n+1; j++) {

   			for(int i=1; i<=n; i++)

   				bin[i]=i;

   			for(int l=1; l<=n+1; l++) {

   				if(l==j)

   					continue;

   				Union(a[l][0],a[l][1]);

   			}

   			t=0;

   			for(int l=1; l<=n; l++)

   				if(bin[l]==l)

   					t++;

   			if(t==1)

   				ans++;

   		}

   		for(int i=1; i<=n+1; i++)

   			for(int j=i+1; j<=n+1; j++) {

   				for(int l=1; l<=n; l++)

   					bin[l]=l;

   				for(int k=1; k<=n+1; k++) {

   					if(k==i||k==j)

   						continue;

   					Union(a[k][0],a[k][1]);

   				}

   				t=0;

   				for(int q=1; q<=n; q++)

   					if(bin[q]==q)

   						t++;

   				if(t==1)

   					ans++;

   			}

   		printf("%d\n",ans);

   	}

   	return 0;
}

  
  
   点击打开链接