NOIP 考前 计算几何练习

Posted 2020-08-19 yyjxx2010xyu

BZOJ 1580

直接解析算出每段的时间然后模拟即可

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <vector>
 7 #include <cmath>
 8 #define Pa pair<double,double>
 9 #define mp make_pair
10 #define pb push_back
11 #define se second
12 #define fi first
13  
14 using namespace std;
15 const int Maxn=50100;
16 const int Inf=0x3f3f3f3f;
17 int n,R,X,Y,Vx,Vy,P[Maxn],Q[Maxn],Vp[Maxn],Vq[Maxn],Ans;
18 vector<Pa> V;
19 inline int Sqr(int x) {return x*x;}
20 inline double Max(double x,double y) {return x>y?x:y;}
21 int main()
22 {
23     // freopen("c.in","r",stdin);
24     scanf("%d%d%d%d%d%d",&n,&R,&X,&Y,&Vx,&Vy);
25     for (int i=1;i<=n;i++)
26     {
27         scanf("%d%d%d%d",&P[i],&Q[i],&Vp[i],&Vq[i]);
28         P[i]-=X,Q[i]-=Y,Vp[i]-=Vx,Vq[i]-=Vy;
29     }
30     for (int i=1;i<=n;i++)
31         if (Vp[i]==0 && Vq[i]==0)
32         {
33             if (Sqr(P[i])+Sqr(Q[i])<=Sqr(R))
34                 V.pb(mp(0,Inf));
35         } else
36         {
37             double Dis=-(P[i]*Vp[i]+Q[i]*Vq[i])/sqrt((Sqr(Vp[i])+Sqr(Vq[i])));
38             if (R*R>=(double)Sqr(P[i])+(double)Sqr(Q[i])-Dis*Dis)
39             {
40                 double tmp=sqrt(R*R-Sqr(P[i])-Sqr(Q[i])+Dis*Dis);
41                 V.pb(mp(Max(Dis-tmp,0.0)/sqrt(Sqr(Vp[i])+Sqr(Vq[i])),(Dis+tmp)/sqrt(Sqr(Vp[i])+Sqr(Vq[i]))));
42             }
43         }
44          
45     sort(V.begin(),V.end());
46     priority_queue<double> T;
47      
48     for (int i=0;i<V.size();i++)
49     {
50         T.push(-V[i].se);
51         while (!T.empty() && -T.top()<V[i].fi) T.pop();
52         Ans=Max(Ans,T.size());
53     }
54     printf("%d\\n",Ans);
55     return 0;
56 }

BZOJ 1580