[codeforces1205B]Shortest Cycle
time limit per test : 1 second
memory limit per test : 256 megabytes
You are given n n n integer numbers a 1 , a 2 , … , a n a_1,a_2,…,a_n a1,a2,…,an. Consider graph on n n n nodes, in which nodes i , j ( i i, j (i i,j(i≠ j ) j) j) are connected if and only if, a i a_i ai AND a j a_j aj ≠0, where AND denotes the bitwise AND operation.
Find the length of the shortest cycle in this graph or determine that it doesn’t have cycles at all.
Input
The first line contains one integer n ( 1 ≤ n ≤ 1 0 5 ) n(1≤n≤10^5) n(1≤n≤105) — number of numbers.
The second line contains n n n integer numbers a 1 , a 2 , … , a n ( 0 ≤ a i ≤ 1 0 18 ) a_1,a_2,…,a_n (0≤a_i≤10^{18}) a1,a2,…,an(0≤ai≤1018).
Output
If the graph doesn’t have any cycles, output − 1 −1 −1. Else output the length of the shortest cycle.
Examples
Input
4
3 6 28 9
Output
4
Input
5
5 12 9 16 48
Output
3
Input
4
1 2 4 8
Output
-1
Note
In the first example, the shortest cycle is (9,3,6,28).
In the second example, the shortest cycle is (5,12,9).
The graph has no cycles in the third example.
题意：
给定 n n n个数字 a 1 a_1 a1 ~ a n a_n an，在 a i a_i ai, a j a_j aj之间有连边，当且仅当 i i i≠ j j j且 a i a_i ai AND a j a_j aj ≠ 0 0 0，求形成的图的最小环。
题解：
首先枚举每一位出现的次数，如果次数大于2，那么答案就是3
如果没有出现次数大于2的，那么将所有位出现次数为2的位所对应的数字取出（最多120个），然后建图之后用floyd计算最小环。
#include
ll ans=INF;
for(int k=1;k<=cnt;k++){
for(int i=1;i<k;i++){
for(int j=i+1;j<k;j++){
ans=min(ans,roc[i][k]+roc[k][j]+mp[i][j]);
}
}
for(int i=1;i<=cnt;i++){
for(int j=1;j<=cnt;j++){
mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
}
}
}
if(ans==INF)puts("-1");
else printf("%lld\n",ans);
/*
for(int i=1;i<=cnt;i++)cout<
return 0;
}