凸包---HDU 2202

题意：给N个点，求着N个点中选择三个联的最大的三角形面积！

注意精度：不然OJ上面会超时的

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
const double eps = 1e-8;
using namespace std;
struct point
{
    int x,y;
} p[50005],res[50005];
int cross(point p0, point p1, point p2)//计算叉积
{
    return(p1.x- p0.x) * (p2.y- p0.y) - (p1.y- p0.y) * (p2.x- p0.x);
}//顺时针扫描，判断大于0（方向改变）的加入凸包，然后回溯
double  dist(point a,point b)//两点距离
{
    return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool cmp(point  a, point  b)//按照y小到大排序，y相等按照x小到大排序
{
    return(a.y< b.y|| (a.y== b.y&& a.x< b.x));
}
int Graham(int n)//求凸包，返回凸包上顶点的个数
{
    int len, top=1;
    sort(p, p+ n, cmp);
    res[0] = p[0];
    res[1] = p[1];
    for(int i=2; i< n; i++)
    {
        while(top&& cross(res[top], res[top-1], p[i])<=eps)top--;
        res[++ top] = p[i];
    }
    len= top;
    res[++ top] = p[n-2];
    for(int i= n-3; i>=0; i--)
    {
        while(top!= len&& cross(res[top], res[top-1], p[i])<=eps)top--;//注意精度问题，不然会超时的
        res[++ top] = p[i];
    }
    return top;
}
int main()
{
    int n,top;
    while(scanf("%d",&n)!=EOF)
    {
        double MAX=-1;
        for(int i=0; i<n; i++)
            scanf("%d%d", &p[i].x ,&p[i].y);
        top=Graham(n);
        for(int i=0; i<top-2; i++)
            for(int j=i+1; j<top-1; j++)
                for(int k=j+1; k<top; k++)
                {
                      double ss=fabs(cross(res[i],res[j],res[k]));
                      if(ss>MAX) MAX=ss;//暴力利用叉积计算三角形的面积
                }
        printf("%.2f\n",MAX/2.0);
    }
    return 0;
}

另一个版本的凸包

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
const double eps = 1e-8;
using namespace std;
struct point
{
    int x,y;
} p[50005],res[50005];
int cross(point p0, point p1, point p2)//计算叉积
{
    return(p1.x- p0.x) * (p2.y- p0.y) - (p1.y- p0.y) * (p2.x- p0.x);
}//顺时针扫描，判断大于0（方向改变）的加入凸包，然后回溯
double  dist(point a,point b)//两点距离
{
    return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool cmp(point  a, point  b)//极角排序，先用arctan()判断角大小，角度由小到大，然后用距离小到大排序
{
     double t1 = atan2(a.y - p[0].y, a.x - p[0].x);
     double t2 = atan2(b.y - p[0].y, b.x - p[0].x);
     return t1>t2||(t1==t2&&dist(p[0],a)<dist(p[0],b));
}
int Graham(int n)//求凸包，返回凸包上顶点的个数
{
    int len, top,tmp=0;
    for(int i=1;i<n;i++)
        if(p[tmp].y>p[i].y||(p[tmp].y==p[i].y&&p[tmp].x>p[i].x))
            tmp=i;
    swap(p[0],p[tmp]);//找出最左下角的点
    sort(p+1, p+ n,cmp);
    res[0] = p[0];
    res[1] = p[1];
    res[2] =p[2];
    top=2;
    for(int i=2; i<n; i++)
    {
        while(top>=2&& cross(res[top], res[top-1], p[i])<=eps)top--;
        res[++ top] = p[i];
    }//只用一次扫描
    res[++top]=p[n-1];//完了要把最后一个点加上去
    return top;
}
int main()
{
    int n,top;
    while(scanf("%d",&n)!=EOF)
    {
        double MAX=-1;
        for(int i=0; i<n; i++)
            scanf("%d%d", &p[i].x ,&p[i].y);
        top=Graham(n);
        for(int i=0; i<top-2; i++)
            for(int j=i+1; j<top-1; j++)
                for(int k=j+1; k<top; k++)
                {
                      double ss=fabs(cross(res[i],res[j],res[k]));
                      if(ss>MAX) MAX=ss;//暴力利用叉积计算三角形的面积
                }
        printf("%.2f\n",MAX/2.0);
    }
    return 0;
}