【POJ 2079】Triangle

Triangle

Description

Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points.

Input

The input consists of several test cases. The first line of each test case contains an integer n, indicating the number of points on the plane. Each of the following n lines contains two integer xi and yi, indicating the ith points. The last line of the input
is an integer ?1, indicating the end of input, which should not be processed. You may assume that 1 <= n <= 50000 and ?10⁴ <= xi, yi <= 10⁴ for all i = 1 . . . n.

Output

For each test case, print a line containing the maximum area, which contains two digits after the decimal point. You may assume that there is always an answer which is greater than zero.

Sample Input

3
3 4
2 6
2 7
5
2 6
3 9
2 0
8 0
6 5
-1

Sample Output

0.50
27.00

Source

Shanghai 2004 Preliminary

旋转卡壳求凸包上的最大三角形面积。

只要把旋转卡壳变形一下：

之前的旋转卡壳是求一个点及其相邻点的对踵点，现在求一个点的时候枚举其他每一个点和这个点的对踵点即可。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define eps 1e-9
using namespace std;
struct Point
{
	double x,y;
}a[50005];
int n,tail,h[50005];
bool cmp(Point a,Point b)
{
	if (a.x==b.x) return a.y<b.y;
	return a.x<b.x;
}
double Cross(Point a,Point b,Point c)
{
	return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
void Graham()
{
	sort(a+1,a+1+n,cmp);
	tail=0;
	for (int i=1;i<=n;i++)
	{
		while (tail>=2&&Cross(a[h[tail-2]],a[h[tail-1]],a[i])<eps)
			tail--;
		h[tail++]=i;
	}
	int tmp=tail;
	h[tail++]=n-1;
	for (int i=n-2;i;i--)
	{
		while (tail>tmp&&Cross(a[h[tail-2]],a[h[tail-1]],a[i])<eps)
			tail--;
		h[tail++]=i;
	}
	tail--;
}
double Rotating_calipers()
{
	double ans=0.0;
	for (int i=0;i<tail;i++)
	{
		int x=i+1;
		for (int j=i+1;j<=tail;j++)
		{
			while (fabs(Cross(a[h[i]],a[h[j]],a[h[x+1]]))>fabs(Cross(a[h[i]],a[h[j]],a[h[x]])))
				x=(x+1)%tail;
			ans=max(ans,fabs(Cross(a[h[i]],a[h[j]],a[h[x]])));
		}
	}
	return ans/2.000;
}
int main()
{
        while (scanf("%d",&n)!=EOF)
	{
		if (n==-1) break;
		for (int i=1;i<=n;i++)
			scanf("%lf%lf",&a[i].x,&a[i].y);
		Graham();
		printf("%.2lf\n",Rotating_calipers());
	}
	return 0;
}