You 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.
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).
If the graph doesn‘t have any cycles, output −1−1. Else output the length of the shortest cycle.
4 3 6 28 9
4
5 5 12 9 16 48
3
4 1 2 4 8
-1
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\n",ans); return 0; }
Codeforces 1206D 思维+floyed求最小环
原文：https://www.cnblogs.com/thmyl/p/12306502.html