codeforces #340B(容易计算几何 自己想复杂了)

codeforces #340B(简单计算几何 自己想复杂了)
B. Maximal Area Quadrilateraltime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard 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

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.

题目大意：给你n个点，让你从这些点里面寻找4个点，使得四边形的面积最大。当然题目说了没有三个点在一条线上，没有重合的点。当时觉得蛮复杂的就没怎么写。
     解题思路：如果能想到把四边形的面积看做是两个三角形面积之和，那就好办了，枚举四边形的对角线，在两边寻找使得三角形面积最大的点，然后得到最大的四边形的面积。时间复杂度为o(n^3)，比较简单了。具体见代码。
     题目地址：B. Maximal Area Quadrilateral
AC代码：

#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++)  //枚举每条对角线            {               double s1=0,s2=0;               for(k=0;k<n;k++)  //向两边找能够延伸最远的点               {                   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;}

查看更多 下一篇