【BZOJ】【P2338】【HNOI2011】【数矩形】【题解】【乱搞】

来源：互联网发布：模拟退火算法的优缺点编辑：程序博客网时间：2024/06/01 10:45

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2338

n^2枚举对角线

记录长度和中点

矩阵的对角线长度相等,中点相同

暴力就好了

~~似乎有数据可以卡?~~

Code:

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int maxn=1501;struct point{LL x,y;point(LL _x=0,LL _y=0):x(_x),y(_y){}bool operator<(point oth)const{return x<oth.x||(x==oth.x&&y<oth.y);}bool operator==(point oth)const{return x==oth.x&&y==oth.y;};point operator+(point oth)const{return point(x+oth.x,y+oth.y);}point operator-(point oth)const{return point(x-oth.x,y-oth.y);}LL operator*(point oth)const{return x*oth.y-y*oth.x;}}a[maxn];LL sqr(LL x){return x*x;}struct Line{LL len;int i,j;point md;bool operator==(Line oth)const{return len==oth.len&&md==oth.md;}bool operator<(Line oth)const{if(len==oth.len)return md<oth.md;return len<oth.len;}}L[maxn*maxn>>1];int n,m;LL LLabs(LL x){return x<0?-x:x;}int main(){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%lld%lld",&a[i].x,&a[i].y);for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){m++;L[m].i=i;L[m].j=j;L[m].len=sqr(a[i].x-a[j].x)+sqr(a[i].y-a[j].y);L[m].md=a[i]+a[j];}}sort(L+1,L+1+m);LL ans=0;for(int i=1;i<=m;i++)for(int j=i-1;j&&L[i]==L[j];j--)ans=max(ans,LLabs((a[L[i].i]-a[L[j].i])*(a[L[i].i]-a[L[j].j])));cout<<ans<<endl;return 0;}

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