[POJ2079]Triangle 凸包+旋转卡壳计算几何

Triangle

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 10242		Accepted: 3074

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 −10 ⁴ <= xi, yi <= 10 ⁴ 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

Sample Output

0.50
27.00

Source

Shanghai 2004 Preliminary

大致题意：

给你一堆点，让你取这些点中的三个点使这三个点组成的三角形面积最大，求最大面积。

题解：

首先我们能一眼看出，所求的点一定是凸包上的点。我们有一种很暴力的做法就是枚举i,j,k这三个点，时间复杂度为O(n^3)，显然再怎么卡也是能把你卡掉的。这时候我们考虑旋转卡壳。我们发现找到i,j后，i,j在旋转的同时，k也满足旋转卡壳的旋转。于是我们可以枚举一个x，表示以i和i+x为底边。然后套用旋转卡壳模板即可。

AC代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;
const int Maxn=50005;
struct node{
	int x,y;
}a[Maxn],ans[Maxn];
int n,k;
node operator -(const node &a,const node &b){
	node c;
	c.x=a.x-b.x;c.y=a.y-b.y;
	return c;
}
inline int cro(node a,node b){
	return a.x*b.y-b.x*a.y; 
}
bool cmp(node x,node y){
	if(x.y!=y.y)return x.y<y.y;
	return x.x<y.x;
}
void convex(){
	k=0;
	sort(a+1,a+1+n,cmp);
	for(int i=1;i<=n;i++){
		while(k>=2&&cro((a[i]-ans[k-1]),(ans[k]-ans[k-1]))>=0)k--;
		ans[++k]=a[i];
	}
	int lim=k;
	for(int i=n;i>=1;i--){
		while(k>=lim+1&&cro((a[i]-ans[k-1]),(ans[k]-ans[k-1]))>=0)k--;
		ans[++k]=a[i];
	}
	if(n>1)k--;
}
inline int add(int x,int y){
	x+=y;
	x%=k+1;
	if(x==0)x=1;
	return x;
}
inline int mx(int x,int y){
	return x>y?x:y;
}
/*inline int step(int x){
	x++;
	if(x>k)x=1;
	return x;
} */
inline int ab(int x){
	return x>0?x:-x;
}/*
void work(){//一种能快不少的奇技淫巧，然而我不会QAQ 
	int p=2,q=3,ansn=0;
	for(int i=1;i<=n;i++){
		scanf("%d%d",&a[i].x,&a[i].y);
	}
	convex();
	for(int i=1;i<=k;i++){
		while(cro((ans[p]-ans[i]),(ans[step(q)]-ans[i]))>cro((ans[p]-ans[i]),(ans[q]-ans[i])))q=step(q);
		ansn=mx(ansn,cro((ans[p]-ans[i]),(ans[q]-ans[i])));
		while(cro((ans[step(p)]-ans[i]),(ans[q]-ans[i]))>cro((ans[p]-ans[i]),(ans[q]-ans[i])))p=step(p);
		ansn=mx(ansn,cro((ans[p]-ans[i]),(ans[q]-ans[i])));
	}
	printf("%.2lf\n",double(ansn/2.0));
}*/
void work(){
	int q=2,ansn=0;
	for(int i=1;i<=n;i++){
		scanf("%d%d",&a[i].x,&a[i].y);
	}
	convex();
	for(int j=1;j<=k/2+1;j++){
		for(int i=1;i<=k;i++){
			while(cro((ans[add(i,j)]-ans[i]),(ans[add(q,1)]-ans[i]))>cro((ans[add(i,j)]-ans[i]),(ans[q]-ans[i])))q=add(q,1);
			ansn=mx(ansn,cro((ans[add(i,j)]-ans[i]),(ans[q]-ans[i])));
		}
	}
	printf("%.2lf\n",double(ansn/2.0));
}
int main(){
	while(1){
		scanf("%d",&n);
		if(n==-1)break;
		work();
	}
	return 0;
}