简单多边形与圆交面积模板

/*
   求多边形与圆相交的面积
*/

#include <iostream>
#include <cstdio>
#include <cmath>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
#define PI acos(-1.0)
#define EPS 1e-10
using namespace std;

inline int DB(double x) {
    if(x<-EPS) return -1;
    if(x>EPS) return 1;
    return 0;
}


struct point {
    double x,y;

    point() {}
    point(double _x,double _y):x(_x),y(_y) {}

    point operator-(point a) {
        return point(x-a.x,y-a.y);
    }
    point operator+(point a) {
        return point(x+a.x,y+a.y);
    }

    double operator*(point a) {
        return x*a.y-y*a.x;
    }

    point oppose() {
        return point(-x,-y);
    }

    double length() {
        return sqrt(x*x+y*y);
    }

    point adjust(double L) {
        L/=length();
        return point(x*L,y*L);
    }

    point vertical() {
        return point(-y,x);
    }

    double operator^(point a) {
        return x*a.x+y*a.y;
    }
};


struct segment {
    point a,b;

    segment() {}
    segment(point _a,point _b):a(_a),b(_b) {}

    point intersect(segment s) {
        double s1=(s.a-a)*(s.b-a);
        double s2=(s.b-b)*(s.a-b);
        double t=s1+s2;
        s1/=t;
        s2/=t;
        return point(a.x*s2+b.x*s1,a.y*s2+b.y*s1);
    }
    point vertical(point p) {
        point t=(b-a).vertical();
        return intersect(segment(p,p+t));
    }
    int isonsegment(point p) {
        return DB(min(a.x,b.x)-p.x)<=0&&
               DB(max(a.x,b.x)-p.x)>=0&&
               DB(min(a.y,b.y)-p.y)<=0&&
               DB(max(a.y,b.y)-p.y)>=0;
    }
};

struct circle {
    point p;
    double R;
};

circle C;
point p[105];
int n;


double cross_area(point a,point b,circle C) {
    point p=C.p;
    double R=C.R;
    int sgn=DB((b-p)*(a-p));
    double La=(a-p).length(),Lb=(b-p).length();
    int ra=DB(La-R),rb=DB(Lb-R);
    double ang=acos(((b-p)^(a-p))/(La*Lb));
    segment t(a,b);
    point s;
    point h,u,temp;
    double ans,L,d,ang1;

    if(!DB(La)||!DB(Lb)||!sgn) ans=0;
    else if(ra<=0&&rb<=0) ans=fabs((b-p)*(a-p))/2;
    else if(ra>=0&&rb>=0) {
        h=t.vertical(p);
        L=(h-p).length();
        if(!t.isonsegment(h)||DB(L-R)>=0) ans=R*R*(ang/2);
        else {
            ans=R*R*(ang/2);
            ang1=acos(L/R);
            ans-=R*R*ang1;
            ans+=R*sin(ang1)*L;
        }
    } else {
        h=t.vertical(p);
        L=(h-p).length();
        s=b-a;
        d=sqrt(R*R-L*L);
        s=s.adjust(d);
        if(t.isonsegment(h+s)) u=h+s;
        else u=h+s.oppose();
        if(ra==1) temp=a,a=b,b=temp;
        ans=fabs((a-p)*(u-p))/2;
        ang1=acos(((u-p)^(b-p))/((u-p).length()*(b-p).length()));
        ans+=R*R*(ang1/2);
    }
    return ans*sgn;
}

double cal_cross(circle C,point p[],int n) {
    double ans=0;
    int i;
    p[n]=p[0];
    for(i=0; i<n; i++) ans+=cross_area(p[i],p[i+1],C);
    return fabs(ans);
}

double x,y,v,det,t,g,R;

int input() {
    scanf("%lf%lf%lf%lf%lf%lf%lf",&x,&y,&v,&det,&t,&g,&R);
    return x||y||v||det||t||g||R;
}


int main() {
    //freopen("test.in","r",stdin);
    while(input()) {
        scanf("%d",&n);
        int i;
        for(i=0; i<n; i++) scanf("%lf%lf",&p[i].x,&p[i].y);
        det=det/180*PI;
        C.R=R;
        C.p.x=x+v*cos(det)*t;
        C.p.y=y+v*sin(det)*t-0.5*g*t*t;
        double area=cal_cross(C,p,n);
        printf("%.2lf
",area);
    }
    return 0;
}

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