HDU3400 三分套三分

题意

就是给你两条线段AB , CD ，一个人在AB以速度p跑,在CD上以q跑,

在其他地方跑速度是r。问你从A到D最少的时间。

三分AB ，然后再三分CD ，模板题目，这题卡精度 eps不能少

 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <cmath>
 #include <algorithm>
 #include <set>
 #include <iostream>
 #include <map>
 #include <stack>
 #include <string>
 #include <vector>
 #define  pi acos(-1.0)
 #define  eps 1e-9
 #define  fi first
 #define  se second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug         printf("******\n")
 #define  mem(a,b)    memset(a,b,sizeof(a))
 #define  name2str(x) #x
 #define  fuck(x)     cout<<#x" = "<<x<<endl
 #define  f(a)        a*a
 #define  sf(n)       scanf("%d", &n)
 #define  sff(a,b)    scanf("%d %d", &a, &b)
 #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
 #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  pf          printf
 #define  FRE(i,a,b)  for(i = a; i <= b; i++)
 #define  FREE(i,a,b) for(i = a; i >= b; i--)
 #define  FRL(i,a,b)  for(i = a; i < b; i++)+
 #define  FRLL(i,a,b) for(i = a; i > b; i--)
 #define  FIN         freopen("data.txt","r",stdin)
 #define  gcd(a,b)    __gcd(a,b)
 #define  lowbit(x)   x&-x
 using namespace std;
 typedef long long  LL;
 typedef unsigned long long ULL;
 const int mod = 1e9 + ;
 const int maxn = 2e5 + ;
 const int INF = 0x3f3f3f3f;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 int t;
 double P, Q, R, len1, len2;
 struct node {
     double x, y;
 } a, b, c, d, temp1, temp2;
 double dist ( node a, node b ) {
     return sqrt ( eps + ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );
 }
 void caltemp1 ( double cnt ) {
     temp1.x = a.x + ( b.x - a.x ) / len1 * cnt;
     temp1.y = a.y + ( b.y - a.y ) / len1 * cnt;
 }
 void caltemp2 ( double cnt ) {
     temp2.x = d.x + ( c.x - d.x ) / len2 * cnt;
     temp2.y = d.y + ( c.y - d.y ) / len2 * cnt;
 }
 double check ( double cnt ) {
     caltemp1 ( cnt );
     double l = , r = len2, ll, rr, ans1, ans2;
     while ( r - l > eps ) {
         ll = (  * l + r ) / , rr = (  * r + l ) / ;
         caltemp2 ( ll );
         ans1 = dist ( temp1, temp2 ) / R + dist ( d, temp2 ) / Q;
         caltemp2 ( rr );
         ans2 = dist ( temp1, temp2 ) / R + dist ( d, temp2 ) / Q;
         if ( ans1 < ans2 ) r = rr;
         else l = ll;
     }
     return ans1 + dist ( temp1, a ) / P;
 }
 int main()  {
     scanf ( "%d", &t );
     while ( t-- ) {
         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 );
         len1 = dist ( a, b ), len2 = dist ( c, d );
         double l = , r = len1, ll, rr;
         while ( r - l > eps ) {
             ll = (  * l + r ) / , rr = (  * r + l ) / ;
             if ( check ( ll ) < check ( rr ) ) r = rr;
             else l = ll;
         }
         printf ( "%.2f\n", check ( ll ) );
     }
     return ;
 }