PAT甲级 - 1107 Social Clusters (30 分) 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

并查集，father[], course[], isRoot[]，三个数组

course[]记录喜欢相同课程的人，发现冲突，则合并

father[]将喜欢相同课程的人合并。

isRoot[]结合FindFather，记录每个cluster的人数

思想上很简单，实现起来真的不容易，想说爱你不简单

#include<cstdio>
#include<algorithm>
using namespace std;
int n;
const int maxn = 1010;
int father[maxn], course[maxn] = { 0 }; // course[i]表明喜欢course[i]的人
int isRoot[maxn] = { false };

int FindFather(int i) {
	//int z = i;
	while (i != father[i]) {
		i = father[i];
	}
	return i;
}

void Union(int a, int b) {
	int faA = FindFather(a);
	int faB = FindFather(b);
	if (faA != faB) {
		father[faA] = faB;
	}
}

void Init(int n) {
	for (int i = 1; i <= n; i++) {
		father[i] = i;
	}
}

bool cmp(int a, int b) {
	return a > b;
}
int main() {
	scanf("%d", &n);
	int v, k;
	Init(n);
	for (int i = 1; i <= n; i++) {
		scanf("%d:", &k);
		for (int j = 0; j < k; j++) {
			scanf("%d", &v);
			if (course[v] == 0) {
				course[v] = i;
			}
			else {
				Union(i, course[v]);
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		isRoot[FindFather(i)]++; //明确每个所属的团体
	}
	int num = 0;
	for (int i = 1; i <= n; i++) {
		if (isRoot[i] != 0) num++;
	}
	printf("%d\n", num);

	sort(isRoot + 1, isRoot + n + 1, cmp);

	for (int i = 1; i <= num; i++) {
		if (isRoot[i] != 0) {
			if (i != 1)
				printf(" ");
			printf("%d", isRoot[i]);
		}
	}
}