poj 2079 Triangle 凸包+旋转卡壳

题意：

给平面上n个点，求这n个点组成的最大三角形面积。

分析：

旋转卡壳，但要注意和求平面最远点对的区别，最大三角形的边不一定在凸包上的，也贴出以前写的求平面最远点对的poj 2187代码作为对比。

代码：

//poj 2079
//sep9
#include <iostream>  
#include <algorithm>  
using namespace std;  
const int maxN=50012;  
  
struct P  
{  
    int x,y;      
}pnt[maxN],cnt[maxN];  
int n;  
  
int cmp(P a,P b)  
{  
    if(a.y!=b.y)  
        return a.y<b.y;  
    return a.x<b.x;  
}  
  
int cross(P a,P b,P c)  
{  
    int x1=b.x-a.x;  
    int y1=b.y-a.y;  
    int x2=c.x-a.x;  
    int y2=c.y-a.y;  
    return x1*y2-x2*y1;  
}  
int graham()  
{  
    sort(pnt,pnt+n,cmp);  
    int i,pos=1;  
    cnt[0]=pnt[0];  
    cnt[1]=pnt[1];  
    for(i=2;i<n;++i){  
        while(pos>0&&cross(cnt[pos-1],cnt[pos],pnt[i])<=0) --pos;  
        cnt[++pos]=pnt[i];        
    }  
    int bak=pos;  
    for(i=n-2;i>=0;--i){  
        while(pos>bak&&cross(cnt[pos-1],cnt[pos],pnt[i])<=0) --pos;  
        cnt[++pos]=pnt[i];  
    }             
    return pos;  
}  
  
int main()  
{  
    while(scanf("%d",&n)==1&&n!=-1){    
        for(int i=0;i<n;++i)  
            scanf("%d%d",&pnt[i].x,&pnt[i].y);  
		int pos=graham();          
    	int ans=0,p=1,i,j,k;
		for(i=j=k=0;i<pos;++i){
			while(abs(cross(cnt[i],cnt[j],cnt[(k+1)%pos]))>abs(cross(cnt[i],cnt[j],cnt[k])))
				k=(k+1)%pos;
			while(abs(cross(cnt[i],cnt[(j+1)%pos],cnt[k]))>abs(cross(cnt[i],cnt[j],cnt[k])))
				j=(j+1)%pos;	
			ans=max(ans,abs(cross(cnt[i],cnt[j],cnt[k])));			
		}    	
    	printf("%.2lf\n",ans/2.0);
	}  
    return 0;     
}   

//poj 2187
//sep9
#include<iostream>
#include<algorithm>
using namespace std;
struct Point
{
	int x,y;	
}points[50012],bag[50012];

int cross(Point p1,Point p2,Point p0)
{
	return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

int cmp(Point a,Point b)
{
	return a.y==b.y?a.x<b.x:a.y<b.y;
}

int dis(Point a,Point b)
{
	return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}

int getBag(Point p[],int n,Point bag[])
{
	int i,bp=1;
    bag[0]=p[0];
	bag[1]=p[1];
	for(i=2;i<n;++i)
	{
		while(bp&&cross(bag[bp-1],p[i],bag[bp])<=0)
			--bp;
		bag[++bp]=p[i];
	}
	int len=bp;
	bag[++bp]=p[n-2];
	for(i=n-3;i>=0;--i)
	{
		while(bp>len&&cross(bag[bp-1],p[i],bag[bp])<=0)
			--bp;
		bag[++bp]=p[i];
	}	
	return bp;
}

int main()
{
	int p,i,n;
	while(scanf("%d",&n)==1)
	{
		for(i=0;i<n;++i)
			scanf("%d%d",&points[i].x,&points[i].y);
		sort(points,points+n,cmp);
		int dif,ans=0,m=getBag(points,n,bag);
		p=1;
		for(i=0;i<m;++i)
		{
			while(abs(cross(bag[i],bag[i+1],bag[p+1]))>abs(cross(bag[i],bag[i+1],bag[p])))
				p=(p+1)%m;
			ans=max(ans,max(dis(bag[i],bag[p]),dis(bag[i+1],bag[p+1])));			
		}
		printf("%d\n",ans);
	}
}

poj 2079 Triangle 凸包+旋转卡壳

原文：http://blog.csdn.net/sepnine/article/details/46546099