PAT甲级——1107 Social Clusters (并查集)

1107 Social Clusters (30 分)

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：将N个人按照兴趣爱好分类，爱好有交集的归为一个社交团体，求社交团体的个数，然后将团体里面的人数进行降序输出。 

思路：在并查集的操作上稍作修改，用根节点的值进行人数的统计，以每一行的第一个爱好代号作为当前数据的root，将之后的X与root进行union操作，若S[rootX]不等于root且 ≤ 0，意味着当前团体已经有了 | S[rootX] | 人，将S[rootX]的值加到S[root]上再进行合并。注意每一行开头的K:需要用字符串数组处理来获取K的值。

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#define MaxNum 1001
using namespace std;
vector <int> S(MaxNum, 0);
bool cmp(int a, int b) {
	return a > b;
}
int getK(char *c);
void unionSet(int root, int X);
int find(int X);
int main()
{
	int N, K;
	scanf("%d", &N);
	for (int i = 0; i < N; i++) {
		int root, X;
		char c[5];
		scanf("%s%d", c, &X);
		K = getK(c);
		root = find(X);
		S[root]--;//人数+1用负数表示
		for (int j = 1; j < K; j++) {
			scanf("%d", &X);
			unionSet(root, X);	
		}
	}
	vector <int> ans;
	for (int i = 1; i < MaxNum; i++) {
		if (S[i] < 0)
			ans.push_back(abs(S[i]));
	}
	sort(ans.begin(), ans.end(), cmp);
	printf("%d\n", ans.size());
	printf("%d", ans[0]);
	for (int i = 1; i < ans.size(); i++)
		printf(" %d", ans[i]);
	printf("\n");
	return 0;
}
int getK(char *c) {
	int sum = 0;
	for (int i = 0; c[i] != ':'; i++)
		sum = sum * 10 + c[i] - '0';
	return sum;
}
void unionSet(int root, int X) {
	int rootX = find(X);
	if(S[rootX] <=0 && rootX != root){
        S[root] += S[rootX];
        S[rootX] = root;
    }
}
int find(int X) {
	if (S[X] <= 0)
		return X;
	else
		return S[X] = find(S[X]);//递归地压缩路径
}