PAT A1107——并查集

 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 }