每日一题 day51 打卡
Analysis
这道题是用非常恶心的三分套三分做的，有一个技巧是不要枚举坐标，枚举两条线段构成三角形的相似比就好了。
了解思路就还挺好写的（尽管我还调了三天）
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #define int long long 7 #define rep(i,s,e) for(register int i=s;i<=e;++i) 8 #define dwn(i,s,e) for(register int i=s;i>=e;--i) 9 using namespace std; 10 inline int read() 11 { 12 int x=0,f=1; 13 char c=getchar(); 14 while(c<'0'||c>'9') {if(c=='-') f=-1; c=getchar();} 15 while(c>='0'&&c<='9') {x=x*10+c-'0'; c=getchar();} 16 return f*x; 17 } 18 inline void write(int x) 19 { 20 if(x<0) {putchar('-'); x=-x;} 21 if(x>9) write(x/10); 22 putchar(x%10+'0'); 23 } 24 struct node 25 { 26 double x,y; 27 }a,b,c,d; 28 double p,q,r; 29 inline node calc_point(node a,node b,double k) 30 { 31 node point; 32 point.x=(b.x-a.x)*k+a.x; 33 point.y=(b.y-a.y)*k+a.y; 34 return point; 35 } 36 inline double calc_dis(node a,node b) 37 { 38 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 39 } 40 inline double check_two(double k1,double k2) 41 { 42 node point_one=calc_point(a,b,k1),point_two=calc_point(c,d,k2); 43 double dis1=calc_dis(a,point_one)/p,dis2=calc_dis(point_one,point_two)/r,dis3=calc_dis(point_two,d)/q; 44 return dis1+dis2+dis3; 45 } 46 inline double check_one(double k) 47 { 48 double l2=0.0,r2=1.0; 49 while(r2-l2>1e-9) 50 { 51 double mid12=l2+(r2-l2)/3.0,mid22=r2-(r2-l2)/3.0; 52 if(check_two(k,mid12)>check_two(k,mid22)) l2=mid12; 53 else r2=mid22; 54 } 55 return check_two(k,l2); 56 } 57 signed main() 58 { 59 scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y); 60 scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y); 61 scanf("%lf%lf%lf",&p,&q,&r); 62 double l1=0.0,r1=1.0; 63 while(r1-l1>1e-9) 64 { 65 double mid11=l1+(r1-l1)/3.0,mid21=r1-(r1-l1)/3.0; 66 if(check_one(mid11)>check_one(mid21)) l1=mid11; 67 else r1=mid21; 68 } 69 printf("%.2lf",check_one(l1)); 70 return 0; 71 }
请各位大佬斧正（反正我不认识斧正是什么意思）