poj 3675 Telescope（三角剖分求多边形和圆交面积）

Telescope

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2271 Accepted: 673

Description

Updog is watching a plane object with a telescope. The field of vision in the telescope can be described as a circle. The center is at the origin and the radius is R. The object can be seen as a simple polygon of N vertexes. Updog wants to know the area of the part of the object that can be seen in the telescope.

Input

The input will contain several test cases. For each case:
The first line contains only one real number R. 
The second line contains an integer N. The following N lines contain two real numbers x_i and y_i each, which describe the coordinates of a vertex. Two vertexes in adjacent lines are also adjacent on the polygon. 
Constraints: 3 ≤ N ≤50, 0.1 ≤ R ≤1000, -1000 ≤ x_i, y_i ≤ 1000

Output

For each test case, output one real number on a separate line, which is the area of the part that can be seen. The result should be rounded to two decimal places.

Sample Input

1030 2010 0-10 0

Sample Output

144.35

Source

POJ Founder Monthly Contest – 2008.07.27, Updog

题意：求一个圆心为原点半径为r的圆和一个多边形的交面积，保证多边形不退化不自交

题解：根据圆心将多边形划分为n部分的三角形和圆的交，然后分情况计算三角形和圆的交面积。。。

就是一堆解释几何的问题，做完能很好的熟悉关于圆的各种计算。。。不过很麻烦。。。很容易错。。。我做了近5个小时

而三角形和圆交的基本情况在这个博客很详细 http://hi.baidu.com/billdu/item/703ad4e15d819db52f140b0b

#include<stdio.h>#include<algorithm>#include<math.h>#define eps 1e-8using namespace std;struct point{    double x,y;    point(){}    point(double x_,double y_):x(x_),y(y_){}}p[58],tp[2],origin;double r,area;int n;double MIN(double x,double y){ return x<y?x:y; }double MAX(double x,double y){ return x>y?x:y; }double cross(point p1,point p2,point p3){    return (p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y);}double dot(point p1,point p2){    return p1.x*p2.x+p1.y*p2.y;}double dis(point p1,point p2){    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}struct point get_intersect(){    struct point temp=point(tp[0].x-tp[1].x,tp[0].y-tp[1].y);    struct point vec=point(temp.y,-temp.x);    struct point origin2=point(origin.x+vec.x,origin.y+vec.y);    double a1=tp[0].y-tp[1].y;    double b1=tp[1].x-tp[0].x;    double c1=(tp[0].x*tp[1].y-tp[1].x*tp[0].y);    double a2=origin.y-origin2.y;    double b2=origin2.x-origin.x;    double c2=(origin.x*origin2.y-origin2.x*origin.y);    double tmd=a1*b2-a2*b1;    return point((b1*c2-b2*c1)/tmd,(a2*c1-a1*c2)/tmd);};int on_line(point p0,point p1,point p2){    if(p0.x>MAX(p1.x,p2.x)) return 0;    if(p0.x<MIN(p1.x,p2.x)) return 0;    if(p0.y>MAX(p1.y,p2.y)) return 0;    if(p0.y<MIN(p1.y,p2.y)) return 0;    return 1;}double get_area(){    double len0=dis(origin,tp[0]);    double len1=dis(origin,tp[1]);    double angle=acos(dot(tp[0],tp[1])/len0/len1);    struct point inter=get_intersect();    double disinter=dis(inter,origin);    double sgn=cross(origin,tp[0],tp[1])<0?-1:1;    double res=0;    if(angle<eps||fabs(cross(origin,tp[0],tp[1]))<eps) return 0;    else if(len0<r+eps&&len1<r+eps)    {        res=fabs(cross(origin,tp[0],tp[1]))/2.0;    }    else if(len0<r-eps||len1<r-eps)    {        if(len1<r-eps)        {            swap(tp[0],tp[1]);            swap(len0,len1);        }        double dis01=dis(tp[0],tp[1]);        struct point mov=point((tp[1].x-tp[0].x)/dis01,(tp[1].y-tp[0].y)/dis01);        double len=sqrt(r*r-disinter*disinter);        struct point interpoint;        interpoint=point(inter.x+mov.x*len,inter.y+mov.y*len);        double angle0=acos(dot(tp[1],interpoint)/len1/r);        res=r*r*angle0/2.0;        res+=fabs(cross(origin,interpoint,tp[0]))/2.0;    }    else    {        int flag=on_line(inter,tp[0],tp[1]);        res=r*r*angle/2.0;        if(flag&&disinter<r)        {            double tangle=2*acos(disinter/r);            res-=r*r*tangle/2.0;            res+=r*disinter*sin(tangle/2.0);        }    }    return res*sgn;}void solve(){    area=0;    for(int i=1;i<=n;i++)    {        tp[0]=p[i],tp[1]=p[i+1];        area+=get_area();    }}int main(){    //freopen("t.txt","r",stdin);    origin=point(0,0);    while(scanf("%lf%d",&r,&n)>0)    {        for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);        p[n+1]=p[1];        solve();        printf("%.2lf\n",fabs(area));    }    return 0;}