旋转多边形是没有前途的,我们考虑旋转敌人,那么答案就是所有人的可行区间长度之和除以$2\pi$
首先对每个敌人找到那些旋转后会落到多边形上的角度,实际上就是圆和一些线段求交,解方程即可,注意判一下落在多边形端点上的情况
把角度排序,每相邻两个角度构成一个区间,在区间内随便取一个角度,把敌人旋转这个角度,判断敌人是否在多边形内,如果是那么整个区间都是可行的
旋转直接套公式:$\left[\matrix{x'\\y'}\right]=\left[\matrix{\cos\theta&-\sin\theta\\\sin\theta&\cos\theta}\right]\left[\matrix{x\\y}\right]$
判断点是否在多边形内:考虑站在这个点,按顺序望向多边形的每个顶点,如果转了$2\pi$弧度就在多边形里,如果转了$0$弧度就在多边形外
注意精度,算夹角时绝对不要用asin或acos,对判别式判断大小不能用eps
#include<stdio.h> #include<math.h> #include<algorithm> using namespace std; typedef double du; const du eps=1e-9,pi2=2*M_PI; bool equ(du a,du b){return fabs(a-b)<eps;} bool lt(du a,du b){return a-b<-eps;} bool inr(du a,du b,du c){ if(a>c)swap(a,c); return lt(a,b)&<(b,c); } struct point{ du x,y; point(du a=0,du b=0){x=a;y=b;} }; point operator-(point a,point b){return point(a.x-b.x,a.y-b.y);} du dot(point a,point b){return a.x*b.x+a.y*b.y;} du cr(point a,point b){return a.x*b.y-a.y*b.x;} du len(point a){return sqrt(a.x*a.x+a.y*a.y);} du dif(point a,point b){ if(equ(len(a)*len(b),0))return 0; if(equ(cr(a,b),0))return lt(0,dot(a,b))?0:-M_PI; du f=atan2(b.y,b.x)-atan2(a.y,a.x); if(lt(cr(a,b),0)){ if(lt(0,f))f-=pi2; }else if(lt(f,0)) f+=pi2; return f; } void eq2(du a,du b,du c,du&d,du&x1,du&x2){ d=b*b-4*a*c; if(d>=0){ x1=(-b+sqrt(d))/(2*a); x2=(-b-sqrt(d))/(2*a); } } du ang[1010]; int M; void chk(point u,point a,point b){ du x1,x2,r; r=len(u); if(equ(r,len(a)))ang[++M]=dif(u,a); if(equ(a.x,b.x)){ if(lt(r,a.x))return; x1=sqrt(r*r-a.x*a.x); x2=-x1; if(inr(a.y,x1,b.y))ang[++M]=dif(u,point(a.x,x1)); if(inr(a.y,x2,b.y))ang[++M]=dif(u,point(a.x,x2)); }else{ du d,K,B; K=(b.y-a.y)/(b.x-a.x); B=a.y-K*a.x; eq2(K*K+1,2*K*B,B*B-r*r,d,x1,x2); if(d<0)return; if(inr(a.x,x1,b.x))ang[++M]=dif(u,point(x1,K*x1+B)); if(inr(a.x,x2,b.x))ang[++M]=dif(u,point(x2,K*x2+B)); } } point e[210],p[510]; int n,m; #define gao(b,c) if(equ(cr(u-b,u-c),0))return 0;\ t+=dif(b-u,c-u); bool inside(point u){ du t=0; int i; for(i=1;i<m;i++){ gao(p[i],p[i+1]) } gao(p[m],p[1]) return equ(t,pi2); } point rot(point a,du ang){ return point(a.x*cos(ang)-a.y*sin(ang),a.x*sin(ang)+a.y*cos(ang)); } #define chk2(a,b,c) if(inside(rot(a,(b+c)*.5)))ans+=(c-b)/pi2; int main(){ int i,j; du ans; scanf("%d%d",&n,&m); for(i=1;i<=n;i++)scanf("%lf%lf",&e[i].x,&e[i].y); for(i=1;i<=m;i++)scanf("%lf%lf",&p[i].x,&p[i].y); ans=0; for(i=1;i<=n;i++){ M=0; for(j=1;j<m;j++){ chk(e[i],p[j],p[j+1]); } chk(e[i],p[m],p[1]); if(M<2){ if(inside(e[i]))ans+=1; }else{ sort(ang+1,ang+M+1); for(j=1;j<M;j++){ chk2(e[i],ang[j],ang[j+1]) } chk2(e[i],ang[M],ang[1]+pi2) } } printf("%.5lf",ans); }