求图中的连通分量的个数

标题求图中的分量的个数连通
今天翻看LRJ的书籍<<算法竞赛入门经典>>的时候,看见一题十分精彩，其中包含了求图的联通分量的个数。利用了并查集，因为害怕搞忘记了并查集的具体写法，所以记录一下。
原题:
Your task is to divide a number of persons into two teams, in such a way, that:

• everyone belongs to one of the teams;
• every team has at least one member;
• every person in the team knows every other person in his team;
• teams are as close in their sizes as possible.

This task may have many solutions. You are to find and output any solution, or to report that the solution does not exist.

Input

     The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

Output

     For simplicity, all persons are assigned a unique integer identifier from 1 to N.

The first line in the input file contains a single integer number N (2 ≤ N ≤ 100) - the total number of persons to divide into teams, followed by N lines - one line per person in ascending order of their identifiers. Each line contains the list of distinct numbers Aij (1 ≤ Aij ≤ N, Aij ≠ i) separated by spaces. The list represents identifiers of persons that ith person knows. The list is terminated by 0.

#include
#define _for(a,b,i) for(int i=a;i<=b;i++)

using namespace std;

const int INF = 1e3;
int graph[INF][INF];
int f[INF];
int find(int x){
	return f[x]==x?f[x]:f[x]=find(f[x]);
}
int add(int x,int y){
	f[find(x)] = find(y);
	return 1;
}
int main(){
	int n,edge;
	while(cin>>n&&n){
		_for(1,n,i){
			f[i] = i;
		}
		memset(graph,0,sizeof(graph));
		cin>>edge;
		for(int i=0;i<edge;i++){
			int number,number1;
			graph[number][number1] = 1;
			add(number,number1);
		}
		int sum = 0;
		_for(1,n,i){
			if(f[i]==i) sum++;
		}
		cout<<"连通分量的个数:"<<sum<<endl;
	}

	return 0;
}