旋转卡壳法求点集最小外接矩形（面积）并输出四个顶点坐标

BZOJ

1185: [HNOI2007]最小矩形覆盖

Time Limit: 10 Sec   Memory Limit: 162 MBSec   Special Judge
Submit: 430   Solved: 202
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Description

 

#include"stdio.h"
#include"string.h"
#include"math.h"
#define M 50006
#define eps 1e-10
#include"stdlib.h"
#define inf 999999999
typedef struct node
{
    double x,y,dis,cos;
}P;
P p[M],q[M],pp[M];
double min(double a,double b)
{
    return a(*(struct node*)b).dis?1:-1;
    else
        return (*(struct node*)b).cos>(*(struct node*)a).cos?1:-1;

}
double pow(double x)
{
    return x*x;
}
double Len(node p0,node p1)
{
    return pow(p1.x-p0.x)+pow(p1.y-p0.y);
}
double COS(node p0,node p1)
{
    double x1=p1.x-p0.x;
    double y1=p1.y-p0.y;
    double x2=1;
    double y2=0;
    return (x1*x2+y1*y2)/sqrt((x1*x1+y1*y1)*(x2*x2+y2*y2));
}
double cross(node p0,node p1,node p2)
{
    double x1=p1.x-p0.x;
    double y1=p1.y-p0.y;
    double x2=p2.x-p0.x;
    double y2=p2.y-p0.y;
    return x1*y2-x2*y1;
}
double dot(node p0,node p1,node p2)
{
    double x1=p1.x-p0.x;
    double y1=p1.y-p0.y;
    double x2=p2.x-p1.x;
    double y2=p2.y-p1.y;
    return x1*x2+y1*y2;
}
node miss(node q1,double a,double b,node q2)//求两直线交点坐标
{
    node ret;
    double c1=a*q1.y-b*q1.x;
    double c2=-a*q2.x-b*q2.y;
    ret.x=-(b*c1+a*c2)/(a*a+b*b);
    ret.y=(a*c1-b*c2)/(a*a+b*b);
    return ret;
}
int main()
{
    int n,i,j;
    while(scanf("%d",&n)!=-1)
    {
        node start;
        int tep;
        start.x=start.y=inf;
        for(i=0;ip[i].y)
            {
                start=p[i];
                tep=i;
            }
            else if(fabs(start.y-p[i].y)p[i].x)
                {
                    start=p[i];
                    tep=i;
                }
            }
        }
        p[tep].dis=0;
        p[tep].cos=1.0;
        for(i=0;ieps||fabs(p[i].dis-p[(i+1)%n].dis)>eps)
                pp[tt++]=p[i];
        }
        if(tt==0)
        {
            printf("%.5lf\n",0.0);
            for(i=0;i<4;i++)
                printf("%.5lf %.5lf\n",p[0].x,p[0].y);
            continue;
        }
        int flag=0;
        for(i=1;ieps)
                flag++;
        }
        if(!flag)
        {
            printf("%.5lf\n",0.0);
            printf("%.5lf %.5lf\n",pp[0].x,pp[0].y);
            printf("%.5lf %.5lf\n",pp[tt-1].x,pp[tt-1].y);
            printf("%.5lf %.5lf\n",pp[tt-1].x,pp[tt-1].y);
            printf("%.5lf %.5lf\n",pp[0].x,pp[0].y);
            continue;
        }
        q[0]=pp[tt-1];//注意tt；
        q[1]=pp[0];
        q[2]=pp[1];
        int cnt=2;
        for(i=2;icross(q[i],q[(i+1)%cnt],q[j%cnt]))
            {
                j++;
            }
            double high=cross(q[i],q[(i+1)%cnt],q[j%cnt])/w;
            while(dot(q[i],q[(i+1)%cnt],q[(k1+1)%cnt])>dot(q[i],q[(i+1)%cnt],q[(k1)%cnt]))
            {
                k1++;
            }
            if(i==0)
                k2=k1;
            while(dot(q[i],q[(i+1)%cnt],q[(k2+1)%cnt])<=dot(q[i],q[(i+1)%cnt],q[(k2)%cnt]))
            {
                k2++;
            }
            double wide=(dot(q[i],q[(i+1)%cnt],q[(k1)%cnt])-dot(q[i],q[(i+1)%cnt],q[(k2)%cnt]))/w;
            if(S>high*wide)
            {
                S=high*wide;//更新四个切点坐标以及旋转的直线的方向向量
                indx[0]=i;
                indx[1]=k1%cnt;
                indx[2]=j%cnt;
                indx[3]=k2%cnt;
                a[0]=q[(i+1)%cnt].x-q[i].x;
                b[0]=q[(i+1)%cnt].y-q[i].y;
                a[1]=b[0];
                b[1]=-a[0];
                a[2]=a[0];
                b[2]=b[0];
                a[3]=b[0];
                b[3]=-a[0];
            }
        }
        printf("%.5lf\n",S);
        node ret[5];
        start.x=start.y=inf;

        for(i=0;i<4;i++)//先找出左下角的点的坐标然后按照极角排序
        {
            ret[i]=miss(q[indx[i]],a[i],b[i],q[indx[(i+1)%4]]);
            //printf("%.5lf %.5lf\n",ret[i].x,ret[i].y);
            if(start.y>ret[i].y)
            {
                start=ret[i];
                tep=i;
            }
            else if(fabs(start.y-ret[i].y)ret[i].x)
                {
                     start=ret[i];
                tep=i;
                }

            }
        }
        ret[tep].dis=0;
        ret[tep].cos=1.5;
        for(i=0;i<4;i++)
        {
            if(i!=tep)
            {
                ret[i].dis=Len(start,ret[i]);
                ret[i].cos=COS(start,ret[i]);
            }
        }
        qsort(ret,4,sizeof(ret[0]),cmp);
        for(i=0;i<4;i++)
            printf("%.5lf %.5lf\n",ret[i].x,ret[i].y);
    }
    return 0;
}

转载于:https://www.cnblogs.com/mypsq/p/4348242.html