HDU 3400 - Line belt （三分）

Problem Description

In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww's speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane.
How long must he take to travel from A to D?

 

Input

The first line is the case number T.
For each case, there are three lines.
The first line, four integers, the coordinates of A and B: Ax Ay Bx By.
The second line , four integers, the coordinates of C and D:Cx Cy Dx Dy.
The third line, three integers, P Q R.
0<= Ax，Ay，Bx，By，Cx，Cy，Dx，Dy<=1000
1<=P，Q，R<=10

 

Output

The minimum time to travel from A to D, round to two decimals.

 

Sample Input

10 0 0 100100 0 100 1002 2 1

 

Sample Output

136.60

---------------------------------------

题意：

在ab上的速度为p，cd上的速度为q，其他路为r，求出从a到d的最少用时。

思路：

在ab上找到一点e, cd上找到一点f, 使得 ae/p + ef/r + fd/q最小，t1 = ae / p, 线性； t2 = ef/r + fd/ q, 递减到递增函数。分别在ab 和 cd上 使用三分（嵌套三分），找到最佳的e ,f点。

eps = 1e-8. 精度不够会WA。

CODE：

#include <iostream>#include <cstdio>#include <cmath>using namespace std;const double eps = 1e-8;struct Point {    double x, y;};double p, q, r;Point a, b, c, d, e, f;double dis(Point u, Point v){    return sqrt((u.x - v.x)*(u.x - v.x) + (u.y - v.y)*(u.y - v.y));}double cal(double m){    f.x = c.x + (d.x - c.x) * m;    f.y = c.y + (d.y - c.y) * m;    return dis(f, d) / q + dis(f, e) / r;}double road(double m){    e.x = a.x + (b.x - a.x) * m;    e.y = a.y + (b.y - a.y) * m;    double l =0.0, r = 1.0, mid, mmid;    while(r - l > eps) {        mid = (l + r)/2.0;        mmid = (mid+r)/2.0;        if(cal(mid) < cal(mmid)) r = mmid;        else l = mid;    }    return cal(l) + dis(a, e) / p;}void solve(){    double l = 0.0, r = 1.0, mid, mmid;    while(r - l > eps) {        mid = (l + r) / 2.0;        mmid = (r + mid) / 2.0;        if(road(mid) < road(mmid)) r = mmid;        else l = mid;     }     printf("%.2lf\n", road(l));}int main(){//freopen("in", "r", stdin);    int T;    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);        solve();    }    return 0;}