1107 Social Clusters
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 Ki (>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
给定社交网络中所有人的兴趣爱好,请你找到所有社会群集。
输入第一行是人数 n ,接下来的 n 行输入编号从 1~n 所有人的爱好。
社会群集判断标准:只要两个人有共同爱好,就属于同一社会群集,比如编号为 3 和 5 的人有共同爱好,3 和 7 的人也有共同爱好,则 3 、5 和 7 都属于同一社会群集。
思路
具体思路如下:
- 输入每个人的爱好,用一个数组
h
来存储每个爱好含有哪些人。 - 初始化并查集,然后将每个爱好中存储的人进行集合合并。
- 统计每个集合的人数并进行降序排序,然后统计集合的个数。
- 输出统计结果,注意行末不能有空格。
代码
#include<bits/stdc++.h> using namespace std; const int N = 1010; int p[N], cnt[N]; vector<int> h[N]; //并查集查找模板 int find(int x) { if (p[x] != x) p[x] = find(p[x]); return p[x]; } int main() { int n, k; cin >> n; //先整理每个爱好有哪些人 for (int i = 0; i < n; i++) { scanf("%d:", &k); while (k--) { int x; cin >> x; h[x].push_back(i); } } //初始化并查集 for (int i = 0; i < n; i++) p[i] = i; //对每个爱好里的人进行集合合并 for (int i = 1; i <= 1000; i++) for (int j = 1; j < h[i].size(); j++) { int a = h[i][0], b = h[i][j]; a = find(a), b = find(b); p[a] = b; } //统计每个集合人数个数并进行降序排序 for (int i = 0; i < n; i++) cnt[find(i)]++; sort(cnt, cnt + n, greater<int>()); //计算集合个数 int res = 0; while (cnt[res]) res++; //输出统计结果 cout << res << endl << cnt[0]; for (int i = 1; i < res; i++) cout << " " << cnt[i]; cout << endl; return 0; }