398. Friends of Friends
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
Memory limit: 65536 kilobytes
input: standard
output: 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;
}