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：

#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 1010;
int father[N];  //存放父亲结点
int isRoot[N] = {0};    //记录每个结点是否作为某个集合的根结点
int course[N] = {0};
int findFather(int x){  //查找x所在集合的根结点
    int a = x;
    while(x!=father[x]){
        x = father[x];
    }
    //路径压缩
    while(a != father[a]){
        int z = a;
        a = father[a];
        father[z] = x;
    }
    return x;
}

void Union(int a,int b){    //合并a和b所在的集合
    int faA = findFather(a);
    int faB = findFather(b);
    if(faA != faB){
        father[faA] = faB;
    }
}
void init(int n){   //初始化father[i]为i,且flag[i]为false
    for(int i=1;i<=n;i++){
        father[i] = i;
        isRoot[i] = false;
    }
}

bool cmp(int a,int b){  //将isRoot数组从大到小排序
    return a > b;
}

int main(){
int n,k,h;
scanf("%d",&n); //人数
init(n);
for(int i=1;i<=n;i++){  //对每个人
    scanf("%d:",&k); //活动个数
    for(int j=0;j<k;j++){   //对每个活动
        scanf("%d",&h);    //输入i号人喜欢的活动h
        if(course[h] == 0){ //如果活动h第一次有人喜欢
            course[h] = i;  //令i喜欢活动h
        }
        Union(i,findFather(course[h])); //合并
    }
}

for(int i=1;i<=n;i++){
    isRoot[findFather(i)]++;    //i的跟结点是findFzther(i),人数加1
}
int ans = 0;    //记录集合数目
for(int i=1;i<=n;i++){
    if(isRoot[i] != 0){
        ans++;  //只统计isRoot[i]不为0的
    }
}
printf("%d
",ans); //输出集合个数
sort(isRoot+1,isRoot+n+1,cmp);
for(int i=1;i<=ans;i++){    //依次输出每个集合内的人数
    printf("%d",isRoot[i]);
    if(i<ans)printf(" ");
}
return 0;
}