1107 Social Clusters（并查集）

When register on a social network, you are always asked to specify your hobbies in order to find some potential friends with the same hobbies. A social cluster is a set of people who have some of their hobbies in common. You are supposed to find all the clusters.

Input Specification:

Each input file contains one test case. For each test case, the first line contains a positive integer N (≤1000), the total number of people in a social network. Hence the people are numbered from 1 to N. Then N lines follow, each gives the hobby list of a person in the format:

K​i​​: h​i​​[1] h​i​​[2] … h​i​​[K​i​​]

where K​i​​ (>0) is the number of hobbies, and h​i​​[j] is the index of the j-th hobby, which is an integer in [1, 1000].

Output Specification:

For each case, print in one line the total number of clusters in the network. Then in the second line, print the numbers of people in the clusters in non-increasing order. The numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

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

Sample Output:

3
4 3 1

题目大意

给出n个集合，每个集合中有若干整数，将含有相同整数的集合合并，求出最终得到的集合数量以及每个集合中整数的个数。

思路

并查集的应用。设person数组为最初每个数字所对应的集合编号，初始化为0，则在输入数字h时，若person[h]不为0，则将当前集合与person[h]对应的集合合并。设cnt数组为每个集合中整数的个数，合并集合时更新。但要注意，在数字输入完毕后，应重新遍历cnt数组，将不为集合根结点的结点i对应的cnt[i]置为0，方便最后的输出。

AC代码

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const int maxn = 1010;
int person[maxn] = {0}, father[maxn], cnt[maxn] = {0}; 
int n, m, h, num = 0;

int findFather(int x){
	if(father[x] == x) return x;
	else{
		int F = findFather(father[x]); 
		father[x] = F; //路径压缩
		return F;
	}
}
void Union(int a, int b){
	int faA = findFather(a);
	int faB = findFather(b);
	if(faA != faB){
		father[faA] = faB;
		cnt[faB] += cnt[faA];
	}
}
bool cmp(int a, int b){
	return a > b;
}

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		father[i] = i;
		cnt[i] = 1;
	}
	for(int i = 1; i <= n; i++){
		scanf("%d:", &m);
		for(int j = 0; j < m; j++){
			scanf("%d", &h);
			if(person[h] == 0){
				person[h] = i;
			} 
			else{ //该数字已在其他集合出现过 
				Union(person[h], i);
			}
		}
	}
	for(int i = 1; i <= n; i++){
		if(findFather(i)!= i) cnt[i] = 0;
		else num++;
	}
	sort(cnt + 1, cnt + n + 1, cmp);
	printf("%d\n", num);
	for(int i = 1; i <= num; i++){
		printf("%d", cnt[i]);
		if(cnt[i + 1] != 0) printf(" ");
	}
	return 0;
}