D. Shortest Cycle
time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputYou are given nn integer numbers a1,a2,…,ana1,a2,…,an. Consider graph on nn nodes, in which nodes ii, jj (i≠ji≠j) are connected if and only if, aiaiAND aj≠0aj≠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 nn (1≤n≤105)(1≤n≤105) — number of numbers.
The second line contains nn integer numbers a1,a2,…,ana1,a2,…,an (0≤ai≤10180≤ai≤1018).
Output
If the graph doesn't have any cycles, output −1−1. Else output the length of the shortest cycle.
Examples
input
Copy
4 3 6 28 9
output
Copy
4
input
Copy
5 5 12 9 16 48
output
Copy
3
input
Copy
4 1 2 4 8
output
Copy
-1
Note
In the first example, the shortest cycle is (9,3,6,28)(9,3,6,28).
In the second example, the shortest cycle is (5,12,9)(5,12,9).
The graph has no cycles in the third example.
#include<iostream> #include<cstdio> #include<cstring> #define maxn 100010 using namespace std; int n,map[210][210],dis[210][210],ans; long long a[maxn]; int main(){ scanf("%d",&n); int cnt=0; for(int i=1;i<=n;i++){ long long x; scanf("%lld",&x); if(x!=0)a[++cnt]=x; } n=cnt; if(n>=200){ puts("3"); return 0; } for(int i=1;i<=n;i++) for(int j=1;j<=n;j++){ if(i==j)continue; dis[i][j]=map[i][j]=100000000; } for(int i=1;i<=n;i++) for(int j=1;j<=n;j++){ if(i==j)continue; if((a[i]&a[j])!=0) map[i][j]=map[j][i]=dis[i][j]=dis[j][i]=1; } ans=0x7fffffff; for(int k=1;k<=n;k++){ for(int i=1;i<k;i++){ for(int j=i+1;j<k;j++){ ans=min(dis[i][j]+map[i][k]+map[k][j],ans); } } for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]); } } } if(ans>n){ puts("-1"); return 0; } else printf("%d ",ans); return 0; }