Signal Interference -（hdu5130）

最新推荐文章于 2018-05-20 18:58:43 发布

原创最新推荐文章于 2018-05-20 18:58:43 发布 · 500 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#数学

数学同时被 2 个专栏收录

27 篇文章

订阅专栏

几何

8 篇文章

订阅专栏

本文介绍了一种计算在两个国家A国和B国之间的战争中，B国建造的无线电塔对A国无线电塔信号干扰的有效覆盖面积的方法。通过输入A国领土的多边形顶点坐标和特定参数k，利用几何算法确定被干扰区域与A国领土相交的面积。

Two countries A-Land and B-Land are at war. The territory of A-Land is a simple polygon with no more than 500 vertices. For military use, A-Land constructed a radio tower (also written as A), and it's so powerful that the whole country was under its signal. To interfere A-Land's communication, B-Land decided to build another radio tower (also written as B). According to an accurate estimation, for any point P, if the euclidean distance between P and B is no more than k (0.2 ≤ k < 0.8) times of the distance between P and A, then point P is not able to receive clear signals from A, i.e. be interfered. Your task is to calculate the area in A-Land's territory that are under B-Land's interference.

Input

There are no more than 100 test cases in the input.

In each test case, firstly you are given a positive integer N indicating the amount of vertices on A-Land's territory, and an above mentioned real number k, which is rounded to 4 digits after the decimal point.

Then N lines follow. Each line contains two integers x and y (|x|, |y| ≤ 1000), indicating a vertex's coordinate on A's territory, in counterclockwise or clockwise order.

The last two lines of a test case give radio tower A and B's coordinates in the same form as vertexes' coordinates. You can assume that A is not equal to B.

Output

For each test case, firstly output the case number, then output your answer in one line following the format shown in sample. Please note that there is a blank after the ':'.

Your solution will be accepted if its absolute error or relative error is no more than 10-6.

This problem is special judged.

Sample Input

Sample Output

Case 1: 0.2729710441

题意：求所有满足PB <= k*PA 的P所在区域与多边形的交面积。

解题思路就是求出圆和多边形相交的面积，这里求出圆的方程就好了，套用模板；

这个是求出的圆的方程

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
using namespace std;

const double eps=1e-8;
const double PI=acos(-1);
const int maxn=1005;
struct point
{
    double x,y;
    point() {}
    point (double xx,double yy):x(xx),y(yy) {}
    friend point operator-(const point &a,const point &b)
    {
        return point (a.x-b.x,a.y-b.y);
    }
    friend point operator *(const point &a,const double &b)
    {
        return point(a.x*b,a.y*b);
    }
    friend point operator /(const point &a,const double &b)
    {
        return point (a.x/b,a.y/b);
    }
} A,B,res[maxn];
double k,r;
int n;
double sqr(double x)
{
    return x*x;
}
int dcmp(double k)
{
    return k<-eps?-1:k>eps?1:0;
}
double dot(const point &a,const point &b)
{
    return a.x*b.x+a.y*b.y;
}
double cross(const point &a,const point &b)
{
    return a.x*b.y-a.y*b.x;
}
double abs(const point o)
{
    return sqrt(dot(o,o));
}
point crosspt(const point &a,const point &b,const point &p,const point &q)
{
    double a1=cross(b-a,p-a);
    double a2=cross(b-a,q-a);
    return (p*a2-q*a1)/(a2-a1);
}
double mysqrt(double n)
{
    return sqrt(max(0.0,n));
}
double sector_area(const point &a,const point &b)
{
    double theta=atan2(a.y,a.x)-atan2(b.y,b.x);
    while(theta<=0) theta+=2*PI;
    while(theta>2*PI) theta-=2*PI;
    theta=min(theta,2*PI-theta);
    return r*r*theta/2;
}
void circle_cross_line(const point &a,const point &b,const point &o,double r,point ret[],int &num)
{
    double x0=o.x,y0=o.y;
    double x1=a.x,y1=a.y;
    double x2=b.x,y2=b.y;
    double dx=x2-x1,dy=y2-y1;
    double A=dx*dx+dy*dy;
    double B=2*dx*(x1-x0)+2*dy*(y1-y0);
    double C=sqr(x1-x0)+sqr(y1-y0)-sqr(r);
    double delta=B*B-4*A*C;
    num=0;
    if(dcmp(delta)>=0)
    {
        double t1=(-B-mysqrt(delta))/2/A;
        double t2=(-B+mysqrt(delta))/2/A;
        if(dcmp(t1-1)<=0&&dcmp(t1)>=0)
            ret[num++]=point(x1+t1*dx,y1+t1*dy);
        if(dcmp(t2-1)<=0&&dcmp(t2)>=0)
            ret[num++]=point(x1+t2*dx,y1+t2*dy);
    }
}
double calc(const point &a,const point &b)
{
    point p[2];
    int num=0;
    int ina=dcmp(abs(a)-r)<0;
    int inb=dcmp(abs(b)-r)<0;
    if (ina&&inb)
        return fabs(cross(a,b)/2.0);
    circle_cross_line(a,b,point(0,0),r,p,num);
    if(ina)
        return sector_area(b,p[0])+fabs(cross(a,p[0]))/2.0;
    if(inb)
        return sector_area(p[0],a)+fabs(cross(p[0],b))/2.0;
    if(num==2)
        return sector_area(a,p[0])+sector_area(p[1],b)+fabs(cross(p[0],p[1]))/2.0;
    return sector_area(a,b);
}
double area()
{
    double ret=0;
    for(int i=0; i<n; i++)
    {
        int sgn=dcmp(cross(res[i],res[i+1]));
        if(sgn!=0)  ret+=sgn*calc(res[i],res[i+1]);
    }
    return ret;
}
void init(point ret[],int n,double xx,double yy)
{
    for(int i=0; i<n; i++)
    {
        ret[i].x-=xx;
        ret[i].y-=yy;
    }
    ret[n]=ret[0];
}
int main()
{
    int cas=0;
    while(~scanf("%d",&n))
    {
        cas++;
        scanf("%lf",&k);
        for(int i=0; i<n; i++)
            scanf("%lf%lf",&res[i].x,&res[i].y);
        scanf("%lf%lf",&A.x,&A.y);
        scanf("%lf%lf",&B.x,&B.y);
        double E=(2*B.x-2*sqr(k)*A.x)/(1.0-sqr(k));
        double F=(2*B.y-2*sqr(k)*A.y)/(1.0-sqr(k));
        double G=(sqr(k*A.x)+sqr(k*A.y)-sqr(B.x)-sqr(B.y))/(1.0-sqr(k));
        double x0=E/2.0,y0=F/2.0;
        r=sqrt(G+sqr(E)/4.0+sqr(F)/4.0);
        for(int i=0; i<n; i++)
        {
            res[i].x-=x0;
            res[i].y-=y0;
        }
        res[n]=res[0];
        printf("Case %d: %.10lf\n",cas,fabs(area()));
    }
    return 0;
}