PAT A1098 . Insertion or Heap Sort (25)

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

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#include "stdio.h"
#include "math.h"
#include "string.h"
//#include "iostream"
//#include "stdlib.h"
#include "vector"
//#include "set"
//#include "map"
//#include "stack"
#include "queue"
#include "algorithm"
using namespace std;

//typedef long long LL;
const int maxn= 1010;
int father[maxn];//存放父亲结点
int isRoot[maxn] = {0};//记录每个结点是否作为某个集合的根结点
int course[maxn] = {0};//记录喜欢活动h的任意一个人的编号
int findFather(int x){//查找x所在集合的根节点
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isRoot[i]为false
for (int i = 1; i <= n; i++) {
father[i] = i;
isRoot[i] = 0;
}
}
bool cmp(int a, int b){
return a > b;
}
int main(){
int n, k, h;
scanf("%d", &n);
init(n);
for (int i = 1; i <= n; i++) {
scanf("%d:", &k);
for (int j = 0; j < k; j++) {//对每个活动
scanf("%d", &h);
if (course[h] == 0) {//h活动第一次有人喜欢
course[h] = i;//记录那个人的名字
}
Union(i, findFather(course[h]));//合并
}
}
for (int i = 1; i <= n; i++) {
isRoot[findFather(i)]++;
}
int ans = 0;
for (int i = 1; i <= n; i++) {
if (isRoot[i]) {
ans++;
}
}
printf("%d\n", ans);
sort(isRoot + 1, isRoot + n + 1, cmp);
for (int i = 1; i <= ans; i++) {
printf("%d", isRoot[i]);
if (i < ans) {
printf(" ");
}
}
return 0;
}