PAT A1107 Social Clusters (30point(s))

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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:
Ki: hi[1] hi[2] ... hi[Ki] where K​i(>0) is the number of hobbies, and hi[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

题意

第一行给出人数N,接下来N号依次是这N个人的爱好,找出连通集(有共同爱好)个数,并降序打印每个集合的人数。

解答

本题很明显要使用并查集,使用map保存每个人的爱好,对所有爱好编号执行并查集算法。使用cnt保存每个集合的大小,然后对所有人的爱好遍历一次,执行cnt[Find(people[i])]++记录集合大小,最后将cnt的值存储到ans数组中,然后打印结果。

#include<iostream>
#include<unordered_map>
#include<vector>
#include<algorithm>

using namespace std;

unordered_map<int, int> parent;

int Find(int x)
{
	// 如果x未在parent中出现过,要初始化
	if (parent.find(x) == parent.end()) parent[x] = x;
	if (parent[x] == x) return x;
	return parent[x] = Find(parent[x]);
}

void Union(int x, int y)
{
	int p1 = Find(x), p2 = Find(y);
	if (p1 != p2) parent[p1] = p2;
}

int main()
{
	int n, m, hobby, tmp;
	scanf("%d", &n);
	vector<int> people(n), ans;
	unordered_map<int, int> cnt;
	for (int i = 0; i < n; i++) {
		scanf("%d: %d", &m, &tmp);
		people[i] = tmp;
		for (int j = 1; j < m; j++) {
			scanf("%d", &hobby);
			Union(tmp, hobby);
		}
	}
	for (int i = 0; i < n; i++) cnt[Find(people[i])]++;
	for (auto it = cnt.begin(); it != cnt.end(); it++) ans.push_back(it->second);
	sort(ans.begin(), ans.end(), greater<int>());
	printf("%lu\n%d", ans.size(), *ans.begin());
	for (auto it = ans.begin() + 1; it != ans.end(); it++) printf(" %d", *it);
	return 0;
}