HDU 2120--Ice_cream's world I【并查集， 判断环的个数】

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 846    Accepted Submission(s): 493

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塔（塔是从0开始到n-1的）

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。

思路：找出环的个数即是土地的块数，每当一组数根节点相同时它们就可能已经和其他的数构成了环，它们有可能是环的开始和结束

<span style="color:#000000;">#include<cstdio>
int per[1010];
int count=0;
void asd()//重要：把每一元素的根节点都设为它本身；
{
for(int i=0;i<1010;i++)/ 
{
	per[i]=i;
}
}
int find(int x)
 {
 int r,i;
r=x;
while(r!=per[r])
{
	r=per[r];
}
return r;
}
void join(int a,int b)//每一次它们根节点相同时（也就是说a能到b）都能构成一个环； 
{
	int fa,fb;
	fa=find(a);
	fb=find(b);
	if(fa!=fb)
	per[fa]=fb;
	else
	count++;
}
int main()
{
	int n,m,i,a,b;
	while(~scanf("%d",&n))
	{
		asd();
		scanf("%d",&m);
		count=0; 
		while(m--)
		{
				scanf("%d %d",&a,&b);
				join(a,b);
		}
		printf("%d\n",count);
	}
	return 0;
}</span>