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。
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef double db; 4 const db eps=1e-6; 5 const db pi=acos(-1); 6 int sign(db k){ 7 if (k>eps) return 1; else if (k<-eps) return -1; return 0; 8 } 9 int cmp(db k1,db k2){return sign(k1-k2);} 10 int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内 11 struct point{ 12 db x,y; 13 point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};} 14 point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};} 15 point operator * (db k1) const{return (point){x*k1,y*k1};} 16 point operator / (db k1) const{return (point){x/k1,y/k1};} 17 int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;} 18 // 逆时针旋转 19 point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};} 20 bool operator < (const point k1) const{ 21 int a=cmp(x,k1.x); 22 if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1; 23 } 24 db abs(){return sqrt(x*x+y*y);} 25 db dis(point k1){return ((*this)-k1).abs();} 26 }; 27 db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;} 28 db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;} 29 point getLL(point k1,point k2,point k3,point k4){ 30 db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2); 31 } 32 struct line{ 33 point p[2]; 34 line(point k1,point k2){p[0]=k1; p[1]=k2;} 35 point& operator [] (int k){return p[k];} 36 point dir(){return p[1]-p[0];} 37 }; 38 int check(line a,line b){//线段与直线 39 if(cross(a[0]-b[0],a[0]-b[1])*cross(a[1]-b[0],a[1]-b[1])>0 ) return 1; 40 return 0; 41 } 42 int t,n;db l,x,y,k;line cut = {{0,0},{0,0}}; 43 vector<line> v; 44 db dfs(line now,int stp){ 45 db ans = 0; 46 if(stp==n)return ans; 47 if(check(now,cut)) { 48 ans+=now.dir().abs(); 49 ans+=dfs({now[1], now[1] + now.dir() / 4}, stp + 1);//原方向 50 ans+=dfs({now[1], now[1] + now.dir().turn(pi/3) / 4}, stp + 1);//逆时针60 51 ans+=dfs({now[1], now[1] + now.dir().turn(-pi/3) / 4}, stp + 1);//顺时针60 52 }else{//砍了 53 point xx = getLL(now[0],now[1],cut[0],cut[1]); 54 ans+=now[0].dis(xx); 55 } 56 return ans; 57 } 58 int main(){ 59 // printf("%.11f\n",pow(3,15)); 60 // printf("%.11f\n",sin(90)); 61 scanf("%d",&t); 62 while (t--){ 63 v.clear(); 64 scanf("%lf%d%lf%lf%lf",&l,&n,&x,&y,&k); 65 line o={{0,0},{0,l}}; 66 cut = {{x,y},{x+10000,y+k*10000}}; 67 db ans = dfs(o,0); 68 printf("%.6f\n",ans); 69 } 70 } 71 /** 72 1 73 16 2 0 18 0 74 */