POJ2504 Bounding box



Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1370 Accepted: 567

Description

The Archeologists of the Current Millenium (ACM) now and then discover ancient artifacts located at vertices of regular polygons. The moving sand dunes of the desert render the excavations difficult and thus once three vertices of a polygon are discovered there is a need to cover the entire polygon with protective fabric.

Input

Input contains multiple cases. Each case describes one polygon. It starts with an integer n <= 50, the number of vertices in the polygon, followed by three pairs of real numbers giving the x and y coordinates of three vertices of the polygon. The numbers are separated by whitespace. The input ends with a n equal 0, this case should not be processed.

Output

For each line of input, output one line in the format shown below, giving the smallest area of a rectangle which can cover all the vertices of the polygon and whose sides are parallel to the x and y axes.

Sample Input

410.00000 0.000000.00000 -10.00000-10.00000 0.00000622.23086 0.42320-4.87328 11.928221.76914 27.5768023156.71567 -13.63236139.03195 -22.04236137.96925 -11.705170

Sample Output

Polygon 1: 400.000Polygon 2: 1056.172Polygon 3: 397.673

模拟题

题意：样例第一行为一个正多边形的边数，2~4行是正多边形三个顶点的坐标，求一个能把该正多边形完全包裹最小的矩形的面积。

思路：通过正多边形的三个顶点求出该正多边形的外接圆，利用向量旋转公式由其中一个点开始求出正多边形所有点的坐标，能包裹住该正多边形的最小的矩形就是要找的矩形，求面积即可。

向量旋转公式：  xx=(xx1-x)*cos(t)-(yy1-y)*sin(t)+x;           yy=(xx1-x)*sin(t)+(yy1-y)*cos(t)+y

代码如下：

//Memory: 224K//Time: 0MS #include<iostream>#include<cmath>using namespace std;int n;double point[3][2];double centre;double pp[55][2];double circle(){//求正多边形外接圆：两垂直平分线的交点就是圆心。     int i,j;    double x,y,k1,k2,b1,b2;    double pi=3.14159265,angle;    double middle1[2],middle2[2];    //求圆心，分斜率k是否存在是否为0.         //k1为0if(point[0][1]==point[1][1])    {        x=(point[0][0]+point[1][0])/2;        //k2不存在         if(point[0][0]==point[2][0])             y=(point[0][1]+point[2][1])/2;        //k2存在且不为0else        {            middle2[0]=(point[2][0]+point[0][0])/2;            middle2[1]=(point[2][1]+point[0][1])/2;            k2=(-1)/((point[2][1]-point[0][1])/(point[2][0]-point[0][0]));            b2=middle2[1]-k2*middle2[0];            y=k2*x+b2;        }    }    //k1不存在     else if(point[0][0]==point[1][0])    {        y=(point[0][1]+point[1][1])/2;        //k2为0 if(point[0][1]==point[2][1])            x=(point[0][0]+point[2][0])/2;        //k2存在且不为0else        {            middle2[0]=(point[2][0]+point[0][0])/2;            middle2[1]=(point[2][1]+point[0][1])/2;            k2=(-1)/((point[2][1]-point[0][1])/(point[2][0]-point[0][0]));            b2=middle2[1]-k2*middle2[0];            x=(y-b2)/k2;        }    }    //k1存在且不为0    else    {        middle1[0]=(point[1][0]+point[0][0])/2;        middle1[1]=(point[1][1]+point[0][1])/2;        k1=(-1)/((point[1][1]-point[0][1])/(point[1][0]-point[0][0]));        b1=middle1[1]-k1*middle1[0];        //k2 =0 if(point[0][1]==point[2][1])        {            x=(point[0][0]+point[2][0])/2;            y=k1*x+b1;        }        //k2不存在         else if(point[0][0]==point[2][0])        {            y=(point[0][1]+point[2][1])/2;            x=(y-b1)/k1;        }        //k2存在且不为0         else        {            middle2[0]=(point[2][0]+point[0][0])/2;            middle2[1]=(point[2][1]+point[0][1])/2;            k2=(-1)/((point[2][1]-point[0][1])/(point[2][0]-point[0][0]));            b2=middle2[1]-k2*middle2[0];            x=(b2-b1)/(k1-k2);            y=k1*x+b1;        }    }    //点到圆心的距离即半径     double r=sqrt((point[0][0]-x)*(point[0][0]-x)+(point[0][1]-y)*(point[0][1]-y));    angle=(2*pi)/(double)n;    pp[0][0]=point[0][0];    pp[0][1]=point[0][1];    //通过向量旋转公式求出所有顶点     for(i=1;i<n;i++)    {        pp[i][0]=(pp[i-1][0]-x)*cos(angle)-(pp[i-1][1]-y)*sin(angle)+x;        pp[i][1]=(pp[i-1][0]-x)*sin(angle)+(pp[i-1][1]-y)*cos(angle)+y;    }    //找出xmax，xmin，ymax，ymin，确定矩形     double maxx=pp[0][0],minx=pp[0][0],maxy=pp[0][1],miny=pp[0][1],ss;    for(i=0;i<n;i++)    {        if(maxx<pp[i][0]) maxx=pp[i][0];        if(minx>pp[i][0]) minx=pp[i][0];        if(maxy<pp[i][1]) maxy=pp[i][1];        if(miny>pp[i][1]) miny=pp[i][1];    }    //求面积     ss=(maxx-minx)*(maxy-miny);    return ss;//返回面积 }int main(){    int i;    double area;    int num=1;    //多样例输入     while(cin>>n&&n)    {        if(n<3) continue;        for(i=0;i<3;i++)            cin>>point[i][0]>>point[i][1];        if(point[0][0]==point[1][0]&&point[1][0]==point[2][0]) continue;         if(point[0][1]==point[1][1]&&point[1][1]==point[2][1]) continue;        double aa=(point[2][1]-point[0][1])/(point[2][0]-point[0][0]);        double bb=(point[1][1]-point[0][1])/(point[1][0]-point[0][0]);        if(aa==bb) continue;        area=circle();        printf("Polygon %d: %.3lf\n",num++,area);    }     return 0;}