Codeforces Round #198 (Div. 2) B. Maximal Area Quadrilateral

B. Maximal Area Quadrilateral

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Iahub has drawn a set of n points in the cartesian plane which he calls "special points". A quadrilateral is a simple polygon without self-intersections with four sides (also called edges) and four vertices (also called corners). Please note that a quadrilateral doesn't have to be convex. A special quadrilateral is one which has all four vertices in the set of special points. Given the set of special points, please calculate the maximal area of a special quadrilateral.

Input

The first line contains integer n (4 ≤ n ≤ 300). Each of the next n lines contains two integers: xi, yi ( - 1000 ≤ xi, yi ≤ 1000) — the cartesian coordinates of ith special point. It is guaranteed that no three points are on the same line. It is guaranteed that no two points coincide.

Output

Output a single real number — the maximal area of a special quadrilateral. The answer will be considered correct if its absolute or relative error does't exceed 10 - 9.

Sample test(s)

input

5
0 0
0 4
4 0
4 4
2 3

output

16.000000

Note

In the test example we can choose first 4 points to be the vertices of the quadrilateral. They form a square by side 4, so the area is 4·4 = 16.

题 意就是找4个点,求最大的四边形的面积,枚举对角线,我们可以求出上三角形的最大值,和下三角形的最大值,这样,我们就可以得出最大的四边形的面积,用叉积可以得出面积,还可以通过其正负,得出是上三角形,还是下三角形,这样,就可以得到了n^3的算法了!

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define M 350
#define eps 0
#define inf 10000000000
struct node {
    double x,y;
}p[M];
double mul(int i,int j,int k){
    return ((p[k].x-p[i].x)*(p[k].y-p[j].y)-(p[k].y-p[i].y)*(p[k].x-p[j].x))/2.0;
}
int main()
{
    int n,i,j,k;
    while(scanf("%d",&n)!=EOF){
        for(i=0;i<n;i++){
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        double lmax=-inf,rmax=-inf,amax=-inf;
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                if(i==j)
                continue;
                lmax=-inf,rmax=-inf;
                for(k=0;k<n;k++){
                    if(i==k||j==k)
                    continue;
                    double temp=mul(i,j,k);
                    if(temp<eps){
                        lmax=max(lmax,-temp);
                    }
                    else {
                        rmax=max(rmax,temp);
                    }
                }
                amax=max(amax,lmax+rmax);
               // printf("%.6f %.6ffdsf\n",amax,lmax+rmax);
            }
        }
        printf("%.6f\n",amax);
    }
    return 0;
}