Codeforces 1206 D - Shortest Cycle

D - Shortest Cycle
思路：n大于某个值肯定有个三元环，否则floyd找最小环。
代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define double long double
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb emplace_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head
 
const int N = 1e5 + 5;
const int INF = 1e8;
LL a[N];
int mp[205][205], dis[205][205];
int n, cnt;
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]), cnt += a[i]==0;
    sort(a+1, a+1+n, greater<LL>());
    n -= cnt;
    if(n >= 200) {
        printf("3\n");
        return 0;
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if(i == j) continue;
            if(((a[i]&a[j]) != 0)) mp[i][j] = dis[i][j] = 1;
            else mp[i][j] = dis[i][j] = INF;
        }
    }
    int ans = INF;
    for (int k = 1; k <= n; ++k) {
        for (int i = 1; i < k; ++i) {
            for(int j = i+1; j < k; ++j) {
                ans = min(ans, dis[i][j]+mp[j][k]+mp[k][i]);
            }
        }
        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 == INF) printf("-1");
    else printf("%d\n", ans);
    return 0;
}