【POJ 2079】Triangle

Description

Input

Output

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

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;
}

【POJ 2079】Triangle

原文：http://blog.csdn.net/regina8023/article/details/43899535