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;
} 求最远点对的代码：
//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);}
}