sgu 398. Friends of Friends 集合set

398. Friends of Friends

Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes

input: standard
output: standard

Social networks are very popular now. They use different types of relationships to organize individual users in a network. In this problem friendship is used as a method to connect users. For each user you are given the list of his friends. Consider friendship as a symmetric relation, so if user a is a friend of user b then b is a friend of a.

A friend of a friend for a is such a user c that c is not a friend of a, but there is such b that b is a friend of a and c is a friend of b. Obviously c ≠ a.

Your task is to find the list of friends of friends for the given user x.

Input

The first line of the input contains integer numbers N and x (1 ≤ N ≤ 50, 1 ≤ x ≤ N), where N is the total number of users and x is user to be processed. Users in the input are specified by their numbers, integers between 1 and N inclusive. The following N lines describe friends list of each user. The i-th line contains integer di (0 ≤ di ≤ 50) — number of friends of the i-th user. After it there are di distinct integers between 1 and N — friends of the i-th user. The list doesn't contain i. It is guaranteed that if user a is a friend of user b then b is a friend of a.

Output

You should output the number of friends of friends of x in the first line. Second line should contain friends of friends of x printed in the increasing order.

Example(s)

sample input	sample output
4 2 1 2 2 1 3 2 4 2 1 3	1 4

sample input	sample output
4 1 3 4 3 2 3 1 3 4 3 1 2 4 3 1 2 3	0

#include <bits/stdc++.h>

using namespace std;

set<int> s;

int m[55][55];
int v[55];

void solve(){
    int n,x;
    scanf("%d %d",&n,&x);
    for(int i=1;i<=n;i++){
        int a;
        scanf("%d",&a);
        while(a--){
            int b;
            scanf("%d",&b);
            m[i][b]=1;
            if(i==x){
                v[b]=1;
            }
        }
    }

    for(int i=1;i<=n;i++){
        if(m[x][i]==1){
            for(int j=1;j<=n;j++){
                if(m[i][j]==1&&v[j]!=1&&j!=x){
                    s.insert(j);
                }
            }
        }
    }

    printf("%d\n",s.size());

    set<int>::iterator ite;

    for(ite=s.begin();ite!=s.end();++ite){
        printf("%d\n",*ite);
    }




}

int main(){
    solve();
    return 0;
}