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: x_i, y_i ( - 1000 ≤ x_i, y_i ≤ 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

50 00 44 04 42 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.

思路：

将四边形看做两个三角形，枚举两个点，构成一条线段，扫描其它点，找到线段两侧的最远点，这四个点构成的三角形最大。复杂度为O(n^3)，不会TLE。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <stack>#include <map>#include <queue>#include <vector>#include <cmath>#define maxn 305using namespace std;int n,m;double ans;struct Node{    int x,y;}pp[maxn],p1,p2,p3;double caldist(Node tt){    return sqrt(tt.x*tt.x*1.0+tt.y*tt.y);}double Det(Node pp1,Node pp2)   // 计算向量叉积{    return pp1.x*pp2.y-pp2.x*pp1.y;}void solve(){    int i,j,k,dir;    double d1,d2,dd,t;    for(i=1;i<=n;i++)    {        for(j=i+1;j<=n;j++)        {            p1.x=pp[j].x-pp[i].x;            p1.y=pp[j].y-pp[i].y;            d1=d2=0;            for(k=1;k<=n;k++)            {                if(k==i||k==j) continue ;                p2.x=pp[k].x-pp[i].x;                p2.y=pp[k].y-pp[i].y;                dd=Det(p1,p2);                if(dd<0) dir=-1;                else dir=1;                t=fabs(dd/caldist(p1));                if(dir==-1)                {                    if(d1<t) d1=t;                }                else                {                    if(d2<t) d2=t;                }            }            if(d1&&d2) ans=max(ans,0.5*caldist(p1)*(d1+d2)); // 有一边没有找到的话就不要更新了        }    }}int main(){    int i,j;    while(~scanf("%d",&n))    {        for(i=1;i<=n;i++)        {            scanf("%d%d",&pp[i].x,&pp[i].y);        }        ans=0;        solve();        printf("%.10f\n",ans);    }    return 0;}