HDOJ 2120 Ice_cream's world I(并查集判断成环)

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 778    Accepted Submission(s): 455

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

 

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

 

Output

Output the maximum number of ACMers who will be awarded.
One answer one line.

 

Sample Input

   
   
   
   

    
    
    
    8 10 0 1 1 2 1 3 2 4 3 4 0 5 5 6 6 7 3 6 4 7
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3
   
   
   
   

 

题意：在一个王国里，建立很多瞭望塔，瞭望塔之间连接有围墙。给出瞭望塔个数n，各个塔之间的围墙数m。接着给出m组A  B；

表示A  B之间有围墙连接，问这些哨塔，围墙把土地分割成了几个部分。

 

英语渣渣的半天没看懂题意，此题是简单的并查集成环。找出环的个数即是土地的块数。直接上代码：

 

#include<cstdio>
int tree[1010],count=0;
int find(int x)
{
	int r=x;
	while(r!=tree[r])
	   r=tree[r];
	return r;
}

void megre(int a,int b)
{
	int fa,fb;
	fa=find(a);
	fb=find(b);
	if(fa!=fb)
	  tree[fa]=fb;
	else
	  count++;	
}

int main()
{
	int n,m,a,b,i;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		count=0;//注意清零，wa了一次！！！ 
		for(i=0;i<n;i++)
		  tree[i]=i;
		while(m--)
		{
			scanf("%d%d",&a,&b);
			megre(a,b);
		}
		printf("%d\n",count);
	}
	return 0;
}