pat1107 Social Clusters 并查集

1107 Social Clusters (30 分)

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

题目大意：给出每个人喜欢的活动，有相同活动的处于一个社交网络，问有多少个社交网络，并从高到低给出每个社交网络人数。

思路：建立一个hobby的数组，当该hobby[x]第一次出现，则hobby[x] = i，i为对应的人的序号，按照题目所给数据：hobby[2] = 1 ,hobby[7]=1, hobby[10] = 1 ,hobby[4] = 2 ,hobby[5] = 3 此时下一个是hobby[4] != 0,所以此时用union函数合并，变成了father[4 ] =2……依次这样做下去，最后得到2的孩子节点有4、6、8，3的孩子节点有5、7；1是独立的，所以有三个社交网络，依次是4、3、1.

#include 
#include "algorithm"
using namespace std;
const int maxx = 1100;
int n,temp,index,ans=0;
int isroot[maxx],father[maxx],hobby[maxx]; //第一个表示该节点是否是根节点，第二个存放父亲节点，第三个记录喜欢活动的人。
void init(int x){
    for (int i = 1; i <= x; ++i) {
        father[i]=i;
        isroot[i] = 0;
    }
}
int findfather(int x){
    int a = x;
    while (x != father[x]){
        x = father[x];
    }
    while (a != father[a]){
        int z = a;
        a = father[a];
        father[z] = x;
    }
    return x;
}
bool cmp(int x,int y){
    return x>y;
}
void Union(int x,int y){
    int fax = findfather(x);
    int fay = findfather(y);
    if (fax != fay){
        father[fax] = fay;
    }
}
int main() {
    cin>>n;
    init(n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d:",&temp);
        for (int j = 0; j < temp; ++j) {
            scanf("%d",&index);
            if (hobby[index] == 0){
                hobby[index] = i;
            }
            Union(i, findfather(hobby[index]));
        }
    }
    for (int i = 1; i <= n; ++i) {
        isroot[findfather(i)]++;
    }
    sort(isroot+1,isroot+n+1, cmp);
    for (int i = 1; i <= n; ++i) {
        if (isroot[i] != 0) ans++;
    }
    printf("%d\n",ans);
    for (int i =1 ; i <= ans; ++i) {
        if (i==1) printf("%d",isroot[1]);
        else printf(" %d",isroot[i]);
    }

}