HDOJ 2120 Ice_cream's world I（并查集）

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 793    Accepted Submission(s): 464

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
   
   
   
   

 

大意：
两个塔之间有一个墙，就是求得把地图分成了几个区域，所以查看是否成环，如果成环，数量+1

ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 100100
int pri[MAXN];
int v[MAXN];
int num;
int find(int x)
{
	int r=x;
	while(r!=pri[r])
	   r=pri[r];
	int i=x,j;
	while(i!=r)
	{
		j=pri[i];
		pri[i]=r;
		i=j;
	}
	return r;
} 
void connect(int xx,int yy)
{
	int a=find(xx);
	int b=find(yy);
	if(a!=b)
	   pri[a]=b;
	else//如果成环，数量+1
	num++;
}
int main()
{
	int n,m;
	int i,a,b;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;i++)//因为有0号塔的存在，所以从0开始，不要习惯性从1开始
		pri[i]=i;
		num=0;
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&a,&b);
			connect(a,b);
		}
		printf("%d\n",num);
	}
	return 0;
}