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 分析:并查集的简单应用:解决关键是将人属于同一集合的合并起来,本题应该通过相同的爱好合并起来;
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std; vector<int> father(), num();
int cnt;//记录集合数 //sort()从大到小排序比较函数
bool cmp(int a, int b){
return a > b;
} //初始化,每个人属于不同集合
void init(int n){
for(int i = ; i <= n; i++){
father[i] = i;
}
} int GetParent(int x){
if(x == father[x])
return x;
father[x] = GetParent(father[x]);//路径压缩
return father[x];
} void merge(int x, int y){
int fx = GetParent(x);
int fy = GetParent(y);
if(fx != fy){
cnt--;//不属于同一集合,合并之后集合数减一
father[fy] = fx;
}
} int main(){
int n, k, h, i;
int hobby[] = {};
scanf("%d", &n); //调用初始化函数
init(n); cnt = n;//初始化集合数为n //处理输入数据,合并集合
for(i = ; i <= n; i++) {
scanf("%d:", &k);
for(int j = ; j < k; j++) {
scanf("%d", &h);
if(hobby[h] == )
hobby[h] = i;
merge(hobby[h], i);
}
}
for(i = ; i <= n; i++)
//这里一定要注意,要找到i所属集合的根节点,num加一,num[father[i]]++是有问题的,因为
//这棵树深度不一定是2.。。即father[i]不一定是i所属集合的根节点
num[GetParent(i)]++; printf("%d\n", cnt); sort(num.begin(), num.end(), cmp); printf("%d", num[]);
for(i = ; i < cnt ; i++)
printf(" %d", num[i]);
return ;
}
 

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