题意:在一些点中找出面积最大的三角形
思路:求出凸包 在凸包中遍历所有三角形找不最大的
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<algorithm>
#define MAX 50010
using namespace std;
typedef struct point
{
double x,y;
};
point p[MAX],pnt[MAX],res[MAX];
bool multi(point p1,point p2,point p0) //[p0,p1]和[p0,p2]的叉积 保持p1,p2,p0的顺序
{ return (p1.x-p0.x)*(p2.y-p0.y)>=(p2.x-p0.x)*(p1.y-p0.y);}
bool operator < (const point &l, const point &r)
{ return l.y < r.y || (l.y == r.y && l.x < r.x); }
int graham(point pnt[], int n, point res[])
{ int i, len, k = 0, top = 1;
sort(pnt, pnt + n);
if (n == 0) return 0; res[0] = pnt[0];
if (n == 1) return 1; res[1] = pnt[1];
if (n == 2) return 2; res[2] = pnt[2];
for (i = 2; i < n; i++)
{ while (top && multi(pnt[i], res[top], res[top-1])) top--;
res[++top] = pnt[i]; //cout<<res[top].x<<"dddd"<<endl;
}
len = top; res[++top] = pnt[n - 2];
for (i = n - 3; i >= 0; i--)
{ while (top!=len && multi(pnt[i], res[top], res[top-1])) top--;
res[++top] = pnt[i];
}
return top; // 返回凸包中点的个数
}
double Area(double a,double b,double c)
{
double p=(a+b+c)*1.0/2;
return sqrt(p*(p-a)*(p-b)*(p-c));
}
int main()
{
int n,top,i,j,k;
while(cin>>n&&n!=-1)
{ double max=0;
for(i=0;i<n;i++)
cin>>pnt[i].x>>pnt[i].y;
top=graham(pnt, n, res);//cout<<top<<endl;
for(i=0;i<top;i++)
{
for(j=i+1;j<top;j++)
{ double a=sqrt((res[i].x-res[j].x)*(res[i].x-res[j].x)+(res[i].y-res[j].y)*(res[i].y-res[j].y));
for(k=j+1;k<top;k++)
{
double b=sqrt((res[i].x-res[k].x)*(res[i].x-res[k].x)+(res[i].y-res[k].y)*(res[i].y-res[k].y));
double c=sqrt((res[j].x-res[k].x)*(res[j].x-res[k].x)+(res[j].y-res[k].y)*(res[j].y-res[k].y));
double s=Area(a,b,c);
if(s>max) max=s;
}
}
}
printf("%.2lf/n",max);
}
// system("pause");
return 0;
}
http://acm.pku.edu.cn/JudgeOnline/problem?id=2079