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:
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
题意：有N个人，每个人喜欢若干项活动，如果两个人有任意一项活动相同，那么就称他们处于同一个社交网络（若A和B属于同一个社交网络，B和C属于同一个社交网络，那么A、B、C属于同一个社交网络）。求这N个人总共形成多少个社交网络。参考代码： 1 #include<cstdio>
2 #include<algorithm>
3 using namespace std;
4 const int N = 1010;
5 int father[N]; //存放父亲结点
6 int isRoot[N] = {0}; //记录每个结点是否作为某个集合的根结点
7 int course[N] = {0};
8 int findFather(int x){ //查找x所在集合的根结点
9 int a = x;
10 while(x!=father[x]){
11 x = father[x];
12 }
13 //路径压缩
14 while(a != father[a]){
15 int z = a;
16 a = father[a];
17 father[z] = x;
18 }
19 return x;
20 }
21
22 void Union(int a,int b){ //合并a和b所在的集合
23 int faA = findFather(a);
24 int faB = findFather(b);
25 if(faA != faB){
26 father[faA] = faB;
27 }
28 }
29 void init(int n){ //初始化father[i]为i,且flag[i]为false
30 for(int i=1;i<=n;i++){
31 father[i] = i;
32 isRoot[i] = false;
33 }
34 }
35
36 bool cmp(int a,int b){ //将isRoot数组从大到小排序
37 return a > b;
38 }
39
40 int main(){
41 int n,k,h;
42 scanf("%d",&n); //人数
43 init(n);
44 for(int i=1;i<=n;i++){ //对每个人
45 scanf("%d:",&k); //活动个数
46 for(int j=0;j<k;j++){ //对每个活动
47 scanf("%d",&h); //输入i号人喜欢的活动h
48 if(course[h] == 0){ //如果活动h第一次有人喜欢
49 course[h] = i; //令i喜欢活动h
50 }
51 Union(i,findFather(course[h])); //合并
52 }
53 }
54
55 for(int i=1;i<=n;i++){
56 isRoot[findFather(i)]++; //i的跟结点是findFzther(i),人数加1
57 }
58 int ans = 0; //记录集合数目
59 for(int i=1;i<=n;i++){
60 if(isRoot[i] != 0){
61 ans++; //只统计isRoot[i]不为0的
62 }
63 }
64 printf("%d\n",ans); //输出集合个数
65 sort(isRoot+1,isRoot+n+1,cmp);
66 for(int i=1;i<=ans;i++){ //依次输出每个集合内的人数
67 printf("%d",isRoot[i]);
68 if(i<ans)printf(" ");
69 }
70 return 0;
71 }
原文地址：https://www.cnblogs.com/mxj961116/p/10586321.html