日常刷题2025-3-15

日常刷题2025-3-15

F - K-th Largest Triplet

满青色

https://atcoder.jp/contests/abc391/tasks/abc391_f

思路：二分答案

一道顶级的二分答案加剪枝优化的题目

清北信息学有题解

代码

#include <bits/stdc++.h>using namespace std;
const int MAXN = 2e5 + 5;
int a[MAXN],b[MAXN],c[MAXN],n,k;long long f(int i,int j,int k){return 1ll*a[i]*b[j] + 1ll*b[j]*c[k] + 1ll*c[k]*a[i];
}bool chk(long long mid){// >=mid的数量是否 >=K int cnt = 0;for(int i = 1;i <= n;++i){if(f(i,1,1) < mid) break;for(int j = 1;j <= n;++j){if(f(i,j,1) < mid) break;for(int k = 1;k <= n;++k){if(f(i,j,k) < mid) break;++cnt;if(cnt == ::k) return true;}}}return false;
}int main(){scanf("%d%d",&n,&k);for(int i = 1;i <= n;++i) scanf("%d",a+i);for(int i = 1;i <= n;++i) scanf("%d",b+i);for(int i = 1;i <= n;++i) scanf("%d",c+i);sort(a+1,a+n+1,[](int x,int y) -> bool {return x > y;});sort(b+1,b+n+1,[](int x,int y) -> bool {return x > y;});sort(c+1,c+n+1,[](int x,int y) -> bool {return x > y;});long long l = 0, r = 3e18,ans = -1;
//	printf("%d\n",chk(34));
//	exit(0);while(l <= r){long long mid = (l + r) >> 1;if(chk(mid)) ans = mid, l = mid+1;else r = mid-1;}printf("%lld\n",ans);return 0;
}