A1107. Social Clusters (30)

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:

83: 2 7 101: 42: 5 31: 41: 31: 44: 6 8 1 51: 4

Sample Output:

34 3 1

#include <cstdio>#include <vector>#include <algorithm>using namespace std;vector<int> father, isRoot;int cmp1(int a, int b){    return a > b;}int findFather(int x) {    int a = x;    while(x != father[x])//原来的x和father[x]不相等，说明x所在不是根节点        x = father[x];    while(a != father[a]) {        int z = a;        a = father[a];        father[z] = x;    }//其实就是father[a]=x;a=father[a];依次向上，将所有的father都写成根节点    return x;}void Union(int a, int b) {    int faA = findFather(a);    int faB = findFather(b);    if(faA != faB)        father[faA] = faB;}int main() {    int n, k, t, cnt = 0;//记录集合数目cnt    int course[1001] = {0};    scanf("%d", &n);    father.resize(n + 1);    isRoot.resize(n + 1);    for(int i = 1; i <= n; i++)        father[i] = i;    for(int i = 1; i <= n; i++) {        scanf("%d:", &k);        for(int j = 0; j < k; j++) {            scanf("%d", &t);            if(course[t] == 0)                course[t] = i;            Union(i, findFather(course[t]));        }    }    for(int i = 1; i <= n; i++)        isRoot[findFather(i)]++;//存的是人数了    for(int i = 1; i <= n; i++) {        if(isRoot[i] != 0)            cnt++;    }    printf("%d\n", cnt);    sort(isRoot.begin(), isRoot.end(), cmp1);    for(int i = 0; i < cnt; i++) {        printf("%d", isRoot[i]);        if(i != cnt - 1) printf(" ");    }    return 0;}