codeforces #340B(簡單計算幾何 自己想複雜了)

果7發表於2013-10-07
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.


題目大意：給你n個點，讓你從這些點裡面尋找4個點，使得四邊形的面積最大。當然題目說了沒有三個點在一條線上，沒有重合的點。當時覺得蠻複雜的就沒怎麼寫。

     解題思路：如果能想到把四邊形的面積看做是兩個三角形面積之和，那就好辦了，列舉四邊形的對角線，在兩邊尋找使得三角形面積最大的點，然後得到最大的四邊形的面積。時間複雜度為o(n^3)，比較簡單了。具體見程式碼。

     題目地址：B. Maximal Area Quadrilateral

AC程式碼：
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
using namespace std;

struct node
{
    double x;
    double y;
};
node p[305];

double multi(node p1,node p2,node p3)  //p1p2向量 叉乘 p1p3向量
{
    double a1,b1,a2,b2;
    a1=p2.x-p1.x,a2=p3.x-p2.x;
    b1=p2.y-p1.y,b2=p3.y-p2.y;
    return a1*b2-a2*b1;
}

int main()
{
    int i,j,k,n;
    double res;
    while(~scanf("%d",&n))
    {
        res=0;
        for(i=0;i<n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        for(i=0;i<n;i++)
            for(j=i+1;j<n;j++)  //列舉每條對角線
            {
               double s1=0,s2=0;
               for(k=0;k<n;k++)  //向兩邊找能夠延伸最遠的點
               {
                   double tmp=multi(p[i],p[j],p[k]);
                   if(tmp>0)
                      s1=max(s1,tmp/2.0);
                   else
                      s2=max(s2,-tmp/2.0);
               }
               if(s1>0&&s2>0&&s1+s2>res)
                  res=s1+s2;
            }
        printf("%f\n",res);
    }
    return 0;
}


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