简单多边形与圆交面积模板

  1 /*
  2    求多边形与圆相交的面积
  3 */
  4 
  5 #include <iostream>
  6 #include <cstdio>
  7 #include <cmath>
  8 #define max(x,y) ((x)>(y)?(x):(y))
  9 #define min(x,y) ((x)<(y)?(x):(y))
 10 #define PI acos(-1.0)
 11 #define EPS 1e-10
 12 using namespace std;
 13 
 14 inline int DB(double x) {
 15     if(x<-EPS) return -1;
 16     if(x>EPS) return 1;
 17     return 0;
 18 }
 19 
 20 
 21 struct point {
 22     double x,y;
 23 
 24     point() {}
 25     point(double _x,double _y):x(_x),y(_y) {}
 26 
 27     point operator-(point a) {
 28         return point(x-a.x,y-a.y);
 29     }
 30     point operator+(point a) {
 31         return point(x+a.x,y+a.y);
 32     }
 33 
 34     double operator*(point a) {
 35         return x*a.y-y*a.x;
 36     }
 37 
 38     point oppose() {
 39         return point(-x,-y);
 40     }
 41 
 42     double length() {
 43         return sqrt(x*x+y*y);
 44     }
 45 
 46     point adjust(double L) {
 47         L/=length();
 48         return point(x*L,y*L);
 49     }
 50 
 51     point vertical() {
 52         return point(-y,x);
 53     }
 54 
 55     double operator^(point a) {
 56         return x*a.x+y*a.y;
 57     }
 58 };
 59 
 60 
 61 struct segment {
 62     point a,b;
 63 
 64     segment() {}
 65     segment(point _a,point _b):a(_a),b(_b) {}
 66 
 67     point intersect(segment s) {
 68         double s1=(s.a-a)*(s.b-a);
 69         double s2=(s.b-b)*(s.a-b);
 70         double t=s1+s2;
 71         s1/=t;
 72         s2/=t;
 73         return point(a.x*s2+b.x*s1,a.y*s2+b.y*s1);
 74     }
 75     point vertical(point p) {
 76         point t=(b-a).vertical();
 77         return intersect(segment(p,p+t));
 78     }
 79     int isonsegment(point p) {
 80         return DB(min(a.x,b.x)-p.x)<=0&&
 81                DB(max(a.x,b.x)-p.x)>=0&&
 82                DB(min(a.y,b.y)-p.y)<=0&&
 83                DB(max(a.y,b.y)-p.y)>=0;
 84     }
 85 };
 86 
 87 struct circle {
 88     point p;
 89     double R;
 90 };
 91 
 92 circle C;
 93 point p[105];
 94 int n;
 95 
 96 
 97 double cross_area(point a,point b,circle C) {
 98     point p=C.p;
 99     double R=C.R;
100     int sgn=DB((b-p)*(a-p));
101     double La=(a-p).length(),Lb=(b-p).length();
102     int ra=DB(La-R),rb=DB(Lb-R);
103     double ang=acos(((b-p)^(a-p))/(La*Lb));
104     segment t(a,b);
105     point s;
106     point h,u,temp;
107     double ans,L,d,ang1;
108 
109     if(!DB(La)||!DB(Lb)||!sgn) ans=0;
110     else if(ra<=0&&rb<=0) ans=fabs((b-p)*(a-p))/2;
111     else if(ra>=0&&rb>=0) {
112         h=t.vertical(p);
113         L=(h-p).length();
114         if(!t.isonsegment(h)||DB(L-R)>=0) ans=R*R*(ang/2);
115         else {
116             ans=R*R*(ang/2);
117             ang1=acos(L/R);
118             ans-=R*R*ang1;
119             ans+=R*sin(ang1)*L;
120         }
121     } else {
122         h=t.vertical(p);
123         L=(h-p).length();
124         s=b-a;
125         d=sqrt(R*R-L*L);
126         s=s.adjust(d);
127         if(t.isonsegment(h+s)) u=h+s;
128         else u=h+s.oppose();
129         if(ra==1) temp=a,a=b,b=temp;
130         ans=fabs((a-p)*(u-p))/2;
131         ang1=acos(((u-p)^(b-p))/((u-p).length()*(b-p).length()));
132         ans+=R*R*(ang1/2);
133     }
134     return ans*sgn;
135 }
136 
137 double cal_cross(circle C,point p[],int n) {
138     double ans=0;
139     int i;
140     p[n]=p[0];
141     for(i=0; i<n; i++) ans+=cross_area(p[i],p[i+1],C);
142     return fabs(ans);
143 }
144 
145 double x,y,v,det,t,g,R;
146 
147 int input() {
148     scanf("%lf%lf%lf%lf%lf%lf%lf",&x,&y,&v,&det,&t,&g,&R);
149     return x||y||v||det||t||g||R;
150 }
151 
152 
153 int main() {
154     //freopen("test.in","r",stdin);
155     while(input()) {
156         scanf("%d",&n);
157         int i;
158         for(i=0; i<n; i++) scanf("%lf%lf",&p[i].x,&p[i].y);
159         det=det/180*PI;
160         C.R=R;
161         C.p.x=x+v*cos(det)*t;
162         C.p.y=y+v*sin(det)*t-0.5*g*t*t;
163         double area=cal_cross(C,p,n);
164         printf("%.2lf
",area);
165     }
166     return 0;
167 }
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转自：http://blog.csdn.net/acm_baihuzi/article/details/48473725
多边形面积求法：https://www.cnblogs.com/xiexinxinlove/p/3708147.html