bzoj-2338 2338: [HNOI2011]数矩形(计算几何)

题目链接：

2338: [HNOI2011]数矩形

Time Limit: 20 Sec  Memory Limit: 128 MB

Description

Input

 

Output

题意：

思路：

求最大的矩形面积,先把这些点转化成线段,记录下线段的长度和中点和两个端点,形成矩形说明对角线长度相等,且共中点,所以把线段按长度和中点排序,如果都相等,然后用三角形的三个顶点坐标计算面积的公式计算最大面积就好了;

AC代码：

/**************************************************************
    Problem: 2338
    User: LittlePointer
    Language: C++
    Result: Accepted
    Time:5172 ms
    Memory:73520 kb
****************************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=4e5+10;
const int maxn=1e3+520;
const double eps=1e-12;

struct PO
{
    LL x,y;
}po[maxn];
struct Seg
{
    PO m;
    int s,e;
    LL dist;
}seg[maxn*maxn];
inline LL dis(int a,int b)
{
    return (po[a].x-po[b].x)*(po[a].x-po[b].x)+(po[a].y-po[b].y)*(po[a].y-po[b].y);
}
int cmp(Seg a,Seg b)
{
    if(a.dist==b.dist)
    {
        if(a.m.x==b.m.x)return a.m.y<b.m.y;
        return a.m.x<b.m.x;
    }
    return a.dist<b.dist;
}
inline LL getans(int a,int b,int c)
{
    LL sum=po[a].x*po[b].y+po[b].x*po[c].y+po[c].x*po[a].y;
    sum=sum-po[a].x*po[c].y-po[b].x*po[a].y-po[c].x*po[b].y;
    if(sum<0)return -sum;
    return sum;
}
inline LL solve(int temp,int f)
{
    return getans(seg[temp].s,seg[temp].e,seg[f].s);
}
int main()
{
    int n;
    read(n);
    For(i,1,n)
    {
        read(po[i].x);
        read(po[i].y);
    }
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            cnt++;
            seg[cnt].s=i;
            seg[cnt].e=j;
            seg[cnt].m.x=po[i].x+po[j].x;
            seg[cnt].m.y=po[i].y+po[j].y;
            seg[cnt].dist=dis(i,j);
        }
    }
    sort(seg,seg+cnt+1,cmp);
    LL ans=0;
    for(int i=1;i<=cnt;i++)
    {
        for(int j=i+1;;j++)
        {
            if(seg[j].dist==seg[i].dist&&seg[j].m.x==seg[i].m.x&&seg[j].m.y==seg[i].m.y)ans=max(ans,solve(j,i));
            else break;
        }
    }
    cout<<ans<<"\n";
    return 0;
}