PAT (Advanced Level)1107. Social Clusters (30)并查集

1107. Social Clusters (30)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

//1107. Social Clusters(30)
#include <iostream>
#include <algorithm>
using namespace std;
bool cmp(int a, int b) {
	return a > b;
}
int f[1001] = { 0 };
int judge[1001] = { 0 };
int h[1001] = { 0 };
int father(int x) {
	if (x != f[x])
		f[x] = father(f[x]);
	return f[x];
}
void Union(int x, int y) {
	int fx = father(x);
	int fy = father(y);
	if (fx != fy)
		f[fx] = fy;
	return;
}
int main() {
	int n, k, s, count = 0;
	cin >> n;
	for (int i = 1; i <= 1001; i++) {
		f[i] = i;
	}
	for (int i = 1; i <= n; i++) {
		scanf("%d:", &k);
		for (int j = 1; j <= k; j++) {
			cin >> s;
			if (h[s] == 0) 
				h[s] = i;
			Union(h[s], i);
		}
	}
	for (int i = 1; i <= n; i++) {
		int x = father(i);
		if (judge[x])
			judge[x]++;
		else {
			judge[x]++;
			count++;
		}
	}
	cout << count << endl;
	sort(judge, judge + n + 1, cmp);
	cout << judge[0];
	for (int i = 1; i <count; i++)
			cout << " "<< judge[i] ;
	return 0;
}