HDU5251 矩形面积 凸包-矩形覆盖

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5251

分析：凸包-矩形覆盖。找出最小的矩形，使其覆盖已知的所有矩形。

实现代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define PR 1e-8
#define N 4010
#define maxdouble 1e20
struct TPoint
{
    double x,y;
};
struct TPolygon
{
    int n;
    TPoint p[N];
} ply;
double MIN(double a,double b)
{
    return a>b?b:a;
}
int dblcmp(double a)
{
    if(fabs(a)<PR) return 0;
    return a>0?1:-1;
}
double dist(TPoint a,TPoint b)//距离
{
    double s1=a.x-b.x;
    double t1=a.y-b.y;
    return sqrt(s1*s1+t1*t1);
}
double cross(TPoint a,TPoint b,TPoint c)//叉积
{
    double s1=b.x-a.x;
    double t1=b.y-a.y;
    double s2=c.x-a.x;
    double t2=c.y-a.y;
    return s1*t2-s2*t1;
}
double dot(TPoint a,TPoint b,TPoint c)//点积
{
    double s1=b.x-a.x;
    double t1=b.y-a.y;
    double s2=c.x-a.x;
    double t2=c.y-a.y;
    return s1*s2+t1*t2;
}
bool cmop(TPoint a,TPoint b)//x、y排序
{
    if(fabs(a.x-b.x)<PR) return a.y<b.y;
    else return a.x<b.x;
}
bool cmp(TPoint a,TPoint b)//叉积内排序
{
    int d1=dblcmp(cross(ply.p[0],a,b));
    return d1>0||(d1==0&&dist(ply.p[0],a)<dist(ply.p[0],b));
}
TPolygon graham()//求凸包
{
    int i,top=2;
    for(i=2; i<ply.n; i++)
    {
        while(top>1&&(dblcmp(cross(ply.p[top-2],ply.p[i],ply.p[top-1])))>=0) top--;
        ply.p[top++]=ply.p[i];
    }
    ply.n=top;
    return ply;
}
double solve()
{
    int i,p=1,q=1,r;
    double minarea=maxdouble,area;
    ply.p[ply.n]=ply.p[0];
    for(i=0; i<ply.n; i++)
    {
        while(dblcmp(cross(ply.p[i],ply.p[i+1],ply.p[p+1])//最上一点
                     -cross(ply.p[i],ply.p[i+1],ply.p[p]))>0)
            p=(p+1)%ply.n;
        while(dblcmp(dot(ply.p[i],ply.p[i+1],ply.p[q+1])//最右一点
                     -dot(ply.p[i],ply.p[i+1],ply.p[q]))>0)
            q=(q+1)%ply.n;
        if(i==0) r=q;
        while(dblcmp(dot(ply.p[i],ply.p[i+1],ply.p[r+1])//最左一点
                     -dot(ply.p[i],ply.p[i+1],ply.p[r]))<=0)
            r=(r+1)%ply.n;
        double d=dist(ply.p[i],ply.p[i+1])*dist(ply.p[i],ply.p[i+1]);
        area=cross(ply.p[i],ply.p[i+1],ply.p[p])*//求面积
             (dot(ply.p[i],ply.p[i+1],ply.p[q])-dot(ply.p[i],ply.p[i+1],ply.p[r]))/d;
        minarea=MIN(area,minarea);//更新
    }
    return minarea;
}
int main()
{
    int t,T=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&ply.n);
        ply.n*=4;
        int i;
        int ans;
        for(i=0; i<ply.n; i++) scanf("%lf%lf",&ply.p[i].x,&ply.p[i].y);
        sort(ply.p,ply.p+ply.n,cmop);//排序
        sort(ply.p+1,ply.p+ply.n,cmp);
        ply=graham();//凸包
        ans=(int)(solve()+0.5);
        printf("Case #%d:\n%d\n",T++,ans);
    }
    return 0;
}