hdoj 2202 最大三角形

题目大意：给定n(3<=n<=50000)个点,求其中任意三个点组成的三角形面积最大,输出该面积.

题目传送：http://acm.hdu.edu.cn/showproblem.php?pid=2202

解题思路:用graham求离散点凸包，随后暴力枚举求最大三角形面积

/*

author:panzg

time:2013/10/22

graham求凸包

*/

#include <iostream>

#include <cmath>

#include <algorithm>

using namespace std;

struct point

{

    double x,y;

};

point p[50010],s[50010];

bool mult(point sp,point ep,point op)

{

    return (sp.x-op.x)*(ep.y-op.y)>=(ep.x-op.x)*(sp.y-op.y);

}

bool  operator  < (const point &l,  const point &r)

{

    return  l.y < r.y || (l.y == r.y && l.x < r.x);

}

int graham(point pnt[],int n,point res[])

{

    int i,len,top=1;

    sort(pnt,pnt+n);

    if(n==0) return 0;

    res[0]=pnt[0];

    if(n==1) return 1;

    res[1]=pnt[1];

    if(n==2) return 2;

    res[2]=pnt[2];

    for(i=2; i<n; i++)

    {

        while(top&&mult(pnt[i],res[top],res[top-1]))

            top--;

        res[++top]=pnt[i];

    }

    len=top;

    res[++top]=pnt[n-2];

    for(i=n-3; i>=0; i--)

    {

        while(top!=len&&mult(pnt[i],res[top],res[top-1])) top--;

        res[++top]=pnt[i];

    }

    return top;

}

double area(point i,point j,point k)

{

    double s=fabs(double(i.x*j.y+j.x*k.y+k.x*i.y-

                         i.y*j.x-j.y*k.x-k.y*i.x)/2);

    return s;

}

double maxx(double x,double y)

{

    return x>y?x:y;

}

int main()

{

    int n,m,i,j,k;



    double ans,t,t1;

    freopen("input.txt","r",stdin);

    while(cin>>n&&n!=0)

    {

        m=0;

        memset(p,0,sizeof(p));

        memset(s,0,sizeof(s));

        for(i=0; i<n; i++)

            cin>>p[i].x>>p[i].y;

        m=graham(p,n,s); //求凸包



        //暴力枚举所有的三角形

        for(ans=0,i=1; i<=m; i++)

            for(j=i+1; j<=m; j++)

                for(k=j+1; k<=m; k++)

                    ans=maxx(ans,area(s[i],s[j],s[k]));

        printf("%.2f\n",ans);

    }

    return 0;

}