NOIP 考前 计算几何练习

BZOJ 1580

直接解析算出每段的时间然后模拟即可

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 #include <vector>
 #include <cmath>
 #define Pa pair<double,double>
 #define mp make_pair
 #define pb push_back
 #define se second
 #define fi first

 using namespace std;
 const int Maxn=;
 const int Inf=0x3f3f3f3f;
 int n,R,X,Y,Vx,Vy,P[Maxn],Q[Maxn],Vp[Maxn],Vq[Maxn],Ans;
 vector<Pa> V;
 inline int Sqr(int x) {return x*x;}
 inline double Max(double x,double y) {return x>y?x:y;}
 int main()
 {
     // freopen("c.in","r",stdin);
     scanf("%d%d%d%d%d%d",&n,&R,&X,&Y,&Vx,&Vy);
     for (int i=;i<=n;i++)
     {
         scanf("%d%d%d%d",&P[i],&Q[i],&Vp[i],&Vq[i]);
         P[i]-=X,Q[i]-=Y,Vp[i]-=Vx,Vq[i]-=Vy;
     }
     for (int i=;i<=n;i++)
         if (Vp[i]== && Vq[i]==)
         {
             if (Sqr(P[i])+Sqr(Q[i])<=Sqr(R))
                 V.pb(mp(,Inf));
         } else
         {
             double Dis=-(P[i]*Vp[i]+Q[i]*Vq[i])/sqrt((Sqr(Vp[i])+Sqr(Vq[i])));
             if (R*R>=(double)Sqr(P[i])+(double)Sqr(Q[i])-Dis*Dis)
             {
                 double tmp=sqrt(R*R-Sqr(P[i])-Sqr(Q[i])+Dis*Dis);
                 V.pb(mp(Max(Dis-tmp,0.0)/sqrt(Sqr(Vp[i])+Sqr(Vq[i])),(Dis+tmp)/sqrt(Sqr(Vp[i])+Sqr(Vq[i]))));
             }
         }

     sort(V.begin(),V.end());
     priority_queue<double> T;

     for (int i=;i<V.size();i++)
     {
         T.push(-V[i].se);
         while (!T.empty() && -T.top()<V[i].fi) T.pop();
         Ans=Max(Ans,T.size());
     }
     printf("%d\n",Ans);
     return ;
 }

BZOJ 1580