[BZOJ]1069: [SCOI2007]最大土地面积

题目大意：给出二维平面上n个点，求最大的由这些点组成的四边形面积。（n<=2000）

思路：求出凸包后旋转卡壳枚举对踵点对作为四边形的对角线，枚举或二分另外两个点，复杂度O(n^2)或O(nlogn)。

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define MN 2000
#define eps 1e-7
struct P{double x,y;}p[MN+5];
P operator-(P a,P b){return (P){a.x-b.x,a.y-b.y};}
double operator*(P a,P b){return a.x*b.y-a.y*b.x;}
double area(P a,P b,P c){return fabs((b-a)*(c-a));}
int dcmp(double x){return fabs(x)<eps?0:x<0?-1:1;}
bool cmp(P a,P b){double c=(a-p[0])*(b-p[0]);return dcmp(c)?dcmp(c)>0:a.y<b.y;}
int n;double ans;
void cal(int l,int r)
{
    double x1=0,x2=0;if(l>r)swap(l,r);int i;
    for(i=l;++i<r;)x1=max(x1,area(p[l],p[r],p[i]));
    for(i=r;++i<=n;)x2=max(x2,area(p[l],p[r],p[i]));
    for(i=0;i<l;++i)x2=max(x2,area(p[l],p[r],p[i]));
    ans=max(ans,x1+x2);
}
int main()
{
    int i,j;
    scanf("%d",&n);
    for(i=j=0;i<n;++i)
    {
        scanf("%lf%lf",&p[i].x,&p[i].y);
        if(p[i].y==p[j].y?p[i].x<p[j].x:p[i].y<p[j].y)j=i;
    }
    swap(p[0],p[j]);sort(p+1,p+n,cmp);
    for(j=1,i=2;i<n;p[++j]=p[i++])
        while(j&&dcmp((p[i]-p[j])*(p[j]-p[j-1]))>=0)--j;
    for(p[n=++j]=p[i=0],j=1;i<n;cal(i,j),cal(++i,j))
        while(area(p[i],p[i+1],p[j+1])>area(p[i],p[i+1],p[j]))++j==n?j=0:0;
    printf("%.3lf",ans/2);
}