洛谷 P2571 [SCOI2010]传送带 题解

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）

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #define int long long
 #define rep(i,s,e) for(register int i=s;i<=e;++i)
 #define dwn(i,s,e) for(register int i=s;i>=e;--i)
 using namespace std;
 inline int read()
 {
     int x=,f=;
     char c=getchar();
     while(c<''||c>'') {if(c=='-') f=-; c=getchar();}
     while(c>=''&&c<='') {x=x*+c-''; c=getchar();}
     return f*x;
 }
 inline void write(int x)
 {
     if(x<) {putchar('-'); x=-x;}
     if(x>) write(x/);
     putchar(x%+'');
 }
 struct node
 {
     double x,y;
 }a,b,c,d;
 double p,q,r;
 inline node calc_point(node a,node b,double k)
 {
     node point;
     point.x=(b.x-a.x)*k+a.x;
     point.y=(b.y-a.y)*k+a.y;
     return point;
 }
 inline double calc_dis(node a,node b)
 {
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 }
 inline double check_two(double k1,double k2)
 {
     node point_one=calc_point(a,b,k1),point_two=calc_point(c,d,k2);
     double dis1=calc_dis(a,point_one)/p,dis2=calc_dis(point_one,point_two)/r,dis3=calc_dis(point_two,d)/q;
     return dis1+dis2+dis3;
 }
 inline double check_one(double k)
 {
     double l2=0.0,r2=1.0;
     while(r2-l2>1e-)
     {
         double mid12=l2+(r2-l2)/3.0,mid22=r2-(r2-l2)/3.0;
         if(check_two(k,mid12)>check_two(k,mid22)) l2=mid12;
         else r2=mid22;
     }
     return check_two(k,l2);
 }
 signed main()
 {
     scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
     scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);
     scanf("%lf%lf%lf",&p,&q,&r);
     double l1=0.0,r1=1.0;
     while(r1-l1>1e-)
     {
         double mid11=l1+(r1-l1)/3.0,mid21=r1-(r1-l1)/3.0;
         if(check_one(mid11)>check_one(mid21)) l1=mid11;
         else r1=mid21;
     }
     printf("%.2lf",check_one(l1));
     return ;
 }

请各位大佬斧正（反正我不认识斧正是什么意思）