19南昌网络赛L

校赛打杂没施展开。

题意：一开始给你一颗 (0,0）到（0,l）的树。

这棵树每一年会长出来三个幼芽(雾)，长度均为l/4,方向分别是左转60，右转60，和不变。

年份<=14

考虑3^14直接暴力存边然后考虑每条边贡献。发现奇难无比。

考虑剪枝。

注意到如果一根树枝的被砍掉了那么他所有的孩子全都不算贡献了。妙啊！变成傻逼题。

抄一下板子。

 #include <bits/stdc++.h>
 using namespace std;
 typedef double db;
 const db eps=1e-;
 const db pi=acos(-);
 int sign(db k){
     if (k>eps) return ; else if (k<-eps) return -; return ;
 }
 int cmp(db k1,db k2){return sign(k1-k2);}
 int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=;}// k3 在 [k1,k2] 内
 struct point{
     db x,y;
     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
     point operator * (db k1) const{return (point){x*k1,y*k1};}
     point operator / (db k1) const{return (point){x/k1,y/k1};}
     int operator == (const point &k1) const{return cmp(x,k1.x)==&&cmp(y,k1.y)==;}
     // 逆时针旋转
     point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
     bool operator < (const point k1) const{
         int a=cmp(x,k1.x);
         if (a==-) return ; else if (a==) return ; else return cmp(y,k1.y)==-;
     }
     db abs(){return sqrt(x*x+y*y);}
     db dis(point k1){return ((*this)-k1).abs();}
 };
 db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
 db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
 point getLL(point k1,point k2,point k3,point k4){
     db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
 }
 struct line{
     point p[];
     line(point k1,point k2){p[]=k1; p[]=k2;}
     point& operator [] (int k){return p[k];}
     point dir(){return p[]-p[];}
 };
 int check(line a,line b){//线段与直线
     if(cross(a[]-b[],a[]-b[])*cross(a[]-b[],a[]-b[])> ) return ;
     return ;
 }
 int t,n;db l,x,y,k;line cut = {{,},{,}};
 vector<line> v;
 db dfs(line now,int stp){
     db ans = ;
     if(stp==n)return ans;
     if(check(now,cut)) {
         ans+=now.dir().abs();
         ans+=dfs({now[], now[] + now.dir() / }, stp + );//原方向
         ans+=dfs({now[], now[] + now.dir().turn(pi/) / }, stp + );//逆时针60
         ans+=dfs({now[], now[] + now.dir().turn(-pi/) / }, stp + );//顺时针60
     }else{//砍了
         point xx = getLL(now[],now[],cut[],cut[]);
         ans+=now[].dis(xx);
     }
     return ans;
 }
 int main(){
 //    printf("%.11f\n",pow(3,15));
 //    printf("%.11f\n",sin(90));
     scanf("%d",&t);
     while (t--){
         v.clear();
         scanf("%lf%d%lf%lf%lf",&l,&n,&x,&y,&k);
         line o={{,},{,l}};
         cut = {{x,y},{x+,y+k*}};
         db ans = dfs(o,);
         printf("%.6f\n",ans);
     }
 }
 /**
 1
 16 2 0 18 0
  */