PAT 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

-----------------------------------这是题目和解题的分割线-----------------------------------

考察并查集的用法，不需要合并爱好，合并人就行了。

#include<cstdio>
#include<algorithm>

using namespace std;

int n,father[1010];

bool cmp(int a,int b)
{
	return a>b;
}

//初始化 
void init(int x)
{
	for(int i=1;i<=n;i++)
		father[i] = i;
}

//寻找x所在集合的根 
int findRoot(int x)
{
	int tmp = x;
	while(x!=father[x]) //找到根结点 
		x = father[x];
	//路径压缩 
	while(tmp!=father[tmp]) //将该路径上所有结点都直接指向根结点x 
	{
		int a = tmp;
		tmp = father[tmp]; //回溯父亲结点 
		father[a] = x; //指向根结点 
	}
	return x; 
}

//合并集合 
void unionS(int a,int b)
{
	int aF = findRoot(a);
	int bF = findRoot(b);
	//如果根不相等，说明不是同一个集合，合并 
	if(aF!=bF) father[aF] = bF;
}

int main()
{
	int k,x,i,j,flag[1010] = {},root[1010] = {},cnt = 0;
	scanf("%d",&n);
	init(n); //初始化，每个人都是一个集合 
	for(i=1;i<=n;i++)
	{
		scanf("%d:",&k);
		for(j=0;j<k;j++)
		{
			scanf("%d",&x);
			//flag[x]用来记录第一次出现x的位置i 
			if(flag[x]==0) flag[x] = i;
			//合并当前位置i和爱好x第一次出现的位置flag[x] 
			unionS(i,findRoot(flag[x]));
		}
	}
	for(i=1;i<=n;i++)
		root[findRoot(i)]++; 
	for(i=1;i<=n;i++)
		if(root[i]) cnt++; //不为空，说明这是一个集合的根 
	sort(root+1,root+1+n,cmp); //排序递减输出 
	printf("%d\n",cnt);
	for(i=1;i<=cnt;i++)
	{
		printf("%d",root[i]);
		if(i!=cnt) printf(" "); //末尾无空格 
	}
	return 0;
}