---

思路

根据所谓的抽屉原理,当n特别大的时候,就会出现几个三元环,这时的答案就是3

其他的跑floyd暴力

---

#include <cstdio> 
 
using namespace std;
 
const int sz = 2000000, maxn = 130, inf = 70000000;
 
char buf[sz], *p1 = buf, *p2 = buf;
inline char gc() { return p1==p2&&(p2=(p1=buf)+fread(buf,1,sz,stdin), p1==p2)?EOF:*p1++; }
inline long long read() {
    long long x = 0;
    int f = 1;
    char ch = gc();
    while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = gc(); }
    while(ch >= '0' && ch <= '9') x = (x<<3)+(x<<1)+(ch&15), ch = gc();
    return x *= f;
}
 
inline int min(int a, int b) { return a < b ? a : b; }
 
long long a[maxn];
int f[maxn][maxn], m[maxn][maxn];
 
int main() {
    int n = (int)read();
    int ans = inf;
    for(int i = 1; i <= n; i++) {
        if(i >= 128) break;
        a[i] = read();
        if(!a[i]) {
            n--; i--;   
        }
    }
    if(n >= 128) {
        printf("3\n");
        return 0;   
    }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            if((a[i] & a[j]) && (i != j))
                f[i][j] = m[i][j] = 1;
            else
                f[i][j] = m[i][j] = inf;
    for(int k = 1; k <= n; k++) {
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++) {
                if(i == j || j == k || k == i) continue;
                ans = min(ans, f[i][j]+m[i][k]+m[k][j]);
            }
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                if(f[i][j] > f[i][k] + f[k][j])
                    f[i][j] = f[i][k] + f[k][j];
    }
    printf("%d\n", ans==inf?-1:ans);    
    return 0;   
}

溜了溜了

原文地址：https://www.cnblogs.com/lyt020321/p/11394068.html