给一个1000个点的多边形（从某个点依次按照外形给出每个节点），这个多边形不一定是凸多边形
再给一个圆，问这个多边形与圆相交区域的周长
我们将这个问题分成两个部分，第一部分是求线段在圆内的长度，第二部分是求圆弧的长度。
对于第一个部分，求线段在圆内的长度，
注意线段两端都在圆外边但是线段与圆有交点的情况。相切的情况可以不去考虑
第二个部分，我们在求第一部分时顺带求出所有线段与圆的交点，依次枚举每段圆弧，我们需要判断这段圆弧是否在多边形内
其实我们只需求出圆弧的中点，对于中点，判断是否在多边形内，射线法或者转角法均可，如果在多边形内就把其长度累加到答案中。
另外特殊判断一下整个圆都在多边形内部的情况
程序是赛后写出来的，还没有提交过_(:зゝ∠)_
#include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<set> #include<map> #include<stack> #include<vector> #include<queue> #include<string> #include<sstream> #define eps 1e-9 #define ALL(x) x.begin(),x.end() #define INS(x) inserter(x,x.begin()) #define rep(i,j,k) for(int i=j;i<=k;i++) #define MAXN 1005 #define MAXM 40005 #define INF 0x3fffffff #define PB push_back #define MP make_pair #define X first #define Y second #define clr(x,y) memset(x,y,sizeof(x)); using namespace std; typedef long long LL; int i,j,k,n,m,x,y,T,big,cas,num,len; bool flag; const double pi=acos(-1.0); int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } struct Vector { double x, y; Vector (double x=0, double y=0) :x(x),y(y) {} Vector operator + (const Vector &B) const { return Vector (x+B.x,y+B.y); } Vector operator - (const Vector &B) const { return Vector(x - B.x, y - B.y); } Vector operator * (const double &p) const { return Vector(x*p, y*p); } Vector operator / (const double &p) const { return Vector(x/p, y/p); } double operator * (const Vector &B) const { return x*B.x + y*B.y;}//点积 double operator ^ (const Vector &B) const { return x*B.y - y*B.x;}//叉积 bool operator < (const Vector &b) const { return x < b.x || (x == b.x && y < b.y); } bool operator ==(const Vector &b) const { return dcmp(x-b.x) == 0 && dcmp(y-b.y) == 0; } }; typedef Vector Point; Point Read(){double x, y;scanf("%lf%lf", &x, &y);return Point(x, y);} double Length(Vector A){ return sqrt(A*A); }//向量的模 double Angle(Vector A, Vector B){return acos(A*B / Length(A) / Length(B)); }//向量的夹角，返回值为弧度 double Area2(Point A, Point B, Point C){ return (B-A)^(C-A); }//向量AB叉乘AC的有向面积 Vector VRotate(Vector A, double rad){return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));}//向量A旋转rad弧度 Point PRotate(Point A, Point B, double rad){return A + VRotate(B-A, rad);}//将B点绕A点旋转rad弧度 Vector Normal(Vector A){double l = Length(A);return Vector(-A.y/l, A.x/l);}//求向量A向左旋转90°的单位法向量,调用前确保A不是零向量 Point GetLineIntersection/*求直线交点，调用前要确保两条直线有唯一交点*/(Point P, Vector v, Point Q, Vector w){double t = (w^(P - Q)) / (v^w);return P + v*t;}//在精度要求极高的情况下，可以自定义分数类 double DistanceToLine/*P点到直线AB的距离*/(Point P, Point A, Point B){Vector v1 = B - A, v2 = P - A;return fabs(v1^v2) / Length(v1);}//不加绝对值是有向距离 double DistanceToSegment/*点到线段的距离*/(Point P, Point A, Point B) { if (A==B) return Length(P-A); Vector v1=B-A,v2=P-A,v3=P-B; if (dcmp(v1*v2)<0) return Length(v2);else if (dcmp(v1*v3)>0) return Length(v3);else return fabs(v1^v2)/Length(v1); } Point GetLineProjection/*点在直线上的射影*/(Point P, Point A, Point B) { Vector v=B-A; return A+v*((v*(P-A))/(v*v)); } bool OnSegment/*判断点是否在线段上(含端点)*/(Point P,Point a1,Point a2) { Vector v1=a1-P,v2=a2-P; if (dcmp(v1^v2)==0 && min(a1.x,a2.x)<=P.x && P.x<=max(a1.x,a2.x) && min(a1.y,a2.y)<=P.y && P.y<=max(a1.y,a2.y)) return true; return false; } bool SegmentInter/*线段相交判定*/(Point a1, Point a2, Point b1, Point b2) { //if (OnSegment(a1,b1,b2) || OnSegment(a2,b1,b2) || OnSegment(b1,a1,a2) || OnSegment(b2,a1,a2)) return 1; //如果只判断线段规范相交（不算交点）,上面那句可以删掉 double c1=(a2-a1)^(b1-a1),c2=(a2-a1)^(b2-a1); double c3=(b2-b1)^(a1-b1),c4=(b2-b1)^(a2-b1); return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0; } bool InTri/*判断点是否在三角形内*/(Point P, Point a,Point b,Point c) { if (dcmp(fabs((c-a)^(c-b))-fabs((P-a)^(P-b))-fabs((P-b)^(P-c))-fabs((P-a)^(P-c)))==0) return true; return false; } double PolygonArea/*求多边形面积，注意凸包P序号从0开始*/(Point *P ,int n) { double ans = 0.0; for(int i=1;i<n-1;i++) ans+=(P[i]-P[0])^(P[i+1]-P[0]); return ans/2; } bool CrossOfSegAndLine/*判断线段是否与直线相交*/(Point a1,Point a2,Point b1,Vector b2) { if (OnSegment(b1,a1,a2) || OnSegment(b1+b2,a1,a2)) return true; return dcmp(b2^(a1-b1))*dcmp(b2^(a2-b1))<0; } Point stand(Point u) { double len=Length(u); return u/len; } int intersCL(Point cen,double R,Point A,Point B,Point &I1,Point &I2)//直线与圆的交点 { double dis=DistanceToLine(cen,A,B); int tt=dcmp(dis-R); if (dcmp(dis-R)>0) return 0; if (dcmp(dis-R)==0) { I1=GetLineProjection(cen,A,B); return 1; } double x=sqrt(R*R-dis*dis); Point p=GetLineProjection(cen,A,B); Vector v=stand(A-B)*x; I1=p+v; I2=p-v; return 2; } Point p[10005],c; double R,ans; bool cmp(Point a,Point b) { return atan2(a.y,a.x)<atan2(b.y,b.x); } //转角发判定点P是否在多边形内部 int isPointInPolygon(Point P, Point* Poly, int n) { int wn=0; for(int i = 0; i < n; ++i) { if(OnSegment(P, Poly[i], Poly[(i+1)%n])) return -1; //在边界上 int k = dcmp((Poly[(i+1)%n] - Poly[i])^( P - Poly[i])); int d1 = dcmp(Poly[i].y - P.y); int d2 = dcmp(Poly[(i+1)%n].y - P.y); if(k > 0 && d1 <= 0 && d2 > 0) wn++; if(k < 0 && d2 <= 0 && d1 > 0) wn--; } if(wn != 0) return 1; //内部 return 0; //外部 } Point middle (Point A,Point B) { return Point((A.x+B.x)/2,(A.y+B.y)/2); } vector <Point> pt; Point print(Point A) { printf("%lf %lf\n",A.x,A.y); } int main() { while (scanf("%d",&n),n) { for (i=0;i<n;i++) p[i]=Read(); c=Read();scanf("%lf",&R); int tot=0; for (i=0;i<n;i++) { Vector v1=p[i]-p[(i-1+n)%n]; Vector v2=p[(i+1)%n]-p[i]; tot+=dcmp(v1^v2); } if (tot<0) { for (i=0;i<n/2;i++) { swap(p[i],p[n-i-1]); } } ans=0; pt.clear(); for (i=0;i<n;i++) { Point A=p[i],B=p[(i+1)%n]; Point I1,I2; int num=intersCL(c,R,A,B,I1,I2); if (num<=1) continue; bool T1=OnSegment(I1,A,B); bool T2=OnSegment(I2,A,B); bool T3=OnSegment(A,I1,I2); bool T4=OnSegment(B,I1,I2); if (A==I1) pt.PB(A-c); if (A==I2) pt.PB(A-c); if (B==I1) pt.PB(B-c); if (B==I2) pt.PB(B-c); if (T1 && T2) { ans+=Length(I2-I1); pt.PB(I1-c);pt.PB(I2-c); }else if (T3 && T4) { ans+=Length(A-B); }else if (!T1 && !T2) { continue; }else if (T1) { if (T4) ans+=Length(I1-B); else ans+=Length(I1-A); pt.PB(I1-c); }else { if (T4) ans+=Length(I2-B); else ans+=Length(I2-A); pt.PB(I2-c); } } sort(pt.begin(),pt.end(),cmp); unique(pt.begin(),pt.end(),cmp); int sz=pt.size(); if (sz==0 && ans==0) { printf("%d\n",(int)round(pi*2*R)); continue; } for (i=0;i<sz;i++) { Point p1=pt[i]; Point p2; if (i==sz-1) { p2=pt[0]; }else p2=pt[i+1]; double a1=atan2(p1.y,p1.x); double a2=atan2(p2.y,p2.x); if (i==sz-1) a2+=2*pi; double a3=(a1+a2)/2; Point pm=VRotate(Point(R,0.0),a3); pm=stand(pm)*R+c; if (isPointInPolygon(pm,p,n)>0) { ans+=(a2-a1)*R; } } printf("%d\n",(int)round(ans)); } return 0; }