首先將速度相減，變成A在動而B不動，若速度為0則顯然永遠不會相交。

枚舉A的每個點以及B的每條線段，計算這三個點共線的時刻。

將時刻排序，對于每個區間進行三分，用半平面交計算相交面積。

注意特判相交面積為0但是存在交點的情況。

時間復雜度$O(n^4\log^2n)$。

?

#include<cstdio> #include<algorithm> #include<cmath> using namespace std; const int N=200; const double eps=1e-9; int sgn(double x){if(x<-eps)return -1;if(x>eps)return 1;return 0; } int n,m,cnt,i,j,vx,vy,x,y;double q[N],ans=-1,anst; struct vec{double x,y;vec(){x=y=0;}vec(double _x,double _y){x=_x,y=_y;}vec operator+(vec v){return vec(x+v.x,y+v.y);}vec operator-(vec v){return vec(x-v.x,y-v.y);}vec operator*(double v){return vec(x*v,y*v);}vec operator/(double v){return vec(x/v,y/v);}double operator*(vec v){return x*v.x+y*v.y;}double len(){return hypot(x,y);}double len_sqr(){return x*x+y*y;} }a[N],b[N],c[N],v,o; double cross(vec a,vec b){return a.x*b.y-a.y*b.x;} bool point_on_segment(vec p,vec a,vec b){return sgn(cross(b-a,p-a))==0&&sgn((p-a)*(p-b))<=0; } int has_intersection(vec a,vec b,vec p,vec q){int d1=sgn(cross(b-a,p-a)),d2=sgn(cross(b-a,q-a)),d3=sgn(cross(q-p,a-p)),d4=sgn(cross(q-p,b-p));if(d1*d2<0&&d3*d4<0)return 1;if(d1==0&&point_on_segment(p,a,b))return -1;if(d2==0&&point_on_segment(q,a,b))return -1;if(d3==0&&point_on_segment(a,p,q))return -1;if(d4==0&&point_on_segment(b,p,q))return -1;return 0; } int line_intersection(vec a,vec b,vec p,vec q,vec&o){double U=cross(p-a,q-p),D=cross(b-a,q-p);if(sgn(D)==0)return 0;o=a+(b-a)*(U/D);return 1; } struct P{double x,y;P(){x=y=0;}P(double _x,double _y){x=_x,y=_y;}P(vec p){x=p.x,y=p.y;}P operator-(const P&a)const{return P(x-a.x,y-a.y);}P operator+(const P&a)const{return P(x+a.x,y+a.y);}P operator*(double a)const{return P(x*a,y*a);} }; namespace Halfplane{ P p[N],a[N]; struct L{P p,v;double a;L(){}L(P _p,P _v){p=_p,v=_v;}bool operator<(const L&b)const{return a<b.a;}void cal(){a=atan2(v.y,v.x);} }line[N],q[N]; int cl; double cross(const P&a,const P&b){return a.x*b.y-a.y*b.x;} void newL(const P&a,const P&b){line[++cl]=L(a,b-a);} bool left(const P&p,const L&l){return cross(l.v,p-l.p)>0;} P pos(const L&a,const L&b){P x=a.p-b.p;double t=cross(b.v,x)/cross(a.v,b.v);return a.p+a.v*t; } double halfplane(){for(int i=1;i<=cl;i++)line[i].cal();sort(line+1,line+cl+1);int h=1,t=1;q[1]=line[1];for(int i=2;i<=cl;i++){while(h<t&&!left(p[t-1],line[i]))t--;while(h<t&&!left(p[h],line[i]))h++;if(fabs(cross(q[t].v,line[i].v))<eps)q[t]=left(q[t].p,line[i])?q[t]:line[i];else q[++t]=line[i];if(h<t)p[t-1]=pos(q[t],q[t-1]);}while(h<t&&!left(p[t-1],q[h]))t--;p[t]=pos(q[t],q[h]);if(t-h<=1)return -1;double ans=0;for(int i=h;i<t;i++)ans+=cross(p[i],p[i+1]);return ans+cross(p[t],p[h]); } } double cal(double T){if(!sgn(T))return -1;double ret=-1;int i,j;for(i=0;i<=n;i++)c[i]=a[i]+(v*T);for(i=0;i<n;i++)for(j=0;j<m;j++)if(has_intersection(c[i],c[i+1],b[j],b[j+1]))ret=0;Halfplane::cl=0;for(i=0;i<n;i++)Halfplane::newL(P(c[i+1]),P(c[i]));for(i=0;i<m;i++)Halfplane::newL(P(b[i+1]),P(b[i]));ret=max(ret,Halfplane::halfplane());if(sgn(ret-ans)>0||(sgn(ret-ans)==0&&T<anst))ans=ret,anst=T;return ret; } int main(){scanf("%d",&n);for(i=0;i<n;i++)scanf("%lf%lf",&a[i].x,&a[i].y);a[n]=a[0];scanf("%d%d",&vx,&vy);scanf("%d",&m);for(i=0;i<m;i++)scanf("%lf%lf",&b[i].x,&b[i].y);b[m]=b[0];scanf("%d%d",&x,&y);vx-=x,vy-=y;if(!vx&&!vy)return puts("never"),0;v=vec(vx,vy);q[cnt=1]=0;for(i=0;i<n;i++)for(j=0;j<m;j++)if(line_intersection(a[i],a[i]+v,b[j],b[j+1],o))q[++cnt]=(o-a[i]).len()/v.len();sort(q+1,q+cnt+1);for(i=1;i<=cnt;i++)cal(q[i]);for(i=1;i<cnt;i++){double l=q[i],r=q[i+1];while(l+1e-6<r){double len=(r-l)/3,m1=l+len,m2=r-len;double f1=cal(m1),f2=cal(m2);if(sgn(f1-f2)>=0)r=m2;else l=m1;}}if(ans<-0.5)puts("never");else printf("%.6f",anst);return 0; }

　　

轉載于:https://www.cnblogs.com/clrs97/p/5652032.html

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