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