hdu 2202 最大三角形 凸包的性质

最大三角形

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4024    Accepted Submission(s): 1439

Problem Description

老师在计算几何这门课上给Eddy布置了一道题目，题目是这样的：给定二维的平面上n个不同的点，要求在这些点里寻找三个点，使他们构成的三角形拥有的面积最大。
Eddy对这道题目百思不得其解，想不通用什么方法来解决，因此他找到了聪明的你，请你帮他解决这个题目。

 

Input

输入数据包含多组测试用例，每个测试用例的第一行包含一个整数n，表示一共有n个互不相同的点，接下来的n行每行包含2个整数xi,yi，表示平面上第i个点的x与y坐标。你可以认为：3 <= n <= 50000 而且 -10000 <= xi, yi <= 10000.

 

Output

对于每一组测试数据，请输出构成的最大的三角形的面积，结果保留两位小数。
每组输出占一行。

 

Sample Input

33 42 63 762 63 92 08 06 67 7

 

Sample Output

1.5027.00

 

Author

Eddy

最大三角形的三个顶点一定在凸包上。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define sqr(x)  ((x)*(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn=50000;const double PI=acos(-1.0);const double eps=1e-10;int dcmp(double x){    if(fabs(x)<eps)  return 0;    else return x<0?-1:1;}struct Point{    double x,y;    Point(double x=0,double y=0):x(x),y(y) {};    bool operator ==(const Point B)const {return dcmp(x-B.x)==0&&dcmp(y-B.y)==0;}    bool operator<(const Point& b)const    {        return dcmp(x-b.x)<0|| dcmp(x-b.x)==0 &&dcmp(y-b.y)<0;    }};typedef Point Vector;Vector operator -(Vector A,Vector B) {return Vector(A.x-B.x,A.y-B.y); }double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}Vector operator +(Vector A,Vector B) {return Vector(A.x+B.x,A.y+B.y); }Vector operator *(Vector A,double p) {return Vector(A.x*p,A.y*p); }Vector operator /(Vector A,double p) {return Vector(A.x/p,A.y/p); }Vector operator -(Vector A)  {return  Vector(-A.x,-A.y);}double Length(Vector A){return sqrt(Dot(A,A));}double torad(double deg){return deg/180*acos(-1.0);}double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}double Area2(Point A,Point B,Point C){return Cross(B-A,C-A);}double Helen(Point A,Point B,Point C)//海伦公式{     Vector AB=B-A;double c= Length(AB);     Vector AC=C-A;double b= Length(AC);     Vector BC=C-B;double a= Length(BC);     double p=(a+b+c)/2;     return sqrt( p*(p-a)*(p-b)*(p-c) );}Point ch[maxn+10];void ConvexHell(Point *p,int n,Point *ch){    sort(p,p+n);     int m=0;    for(int i=0;i<n;i++)    {        while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-1]   )<=0   )  m--;        ch[m++]=p[i];     }    int k=m;    for(int i=n-2;i>=0;i--)    {        while(m>k &&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-1] )<=0)   m--;        ch[m++]=p[i];    }    if(n>1)  m--;    double ans=0;    for0(i,m)    {        for(int j=i+1;j<m;j++)        {            for(int k=j+1;k<m;k++)            {                 ans=max(ans,Helen(ch[i],ch[j],ch[k]  ));            }        }    }    printf("%.2f\n",ans);}int n;Point p[maxn+10];int main(){   std::ios::sync_with_stdio(false);   while(cin>>n)   {       for0(i,n)  cin>>p[i].x>>p[i].y;       ConvexHell(p,n,ch);   }   return 0;}