Line belt （三分查找）

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 629 Accepted Submission(s): 325

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
1
0 0 0 100
100 0 100 100
2 2 1

Sample Output
136.60
/*
自己弄了老一会才弄出来 就是一个典型的三分算法 不过这个题是一个三分套三分算法 首先我们先对ab线段求三分一边求一边对cd线段求三分 然后算出时间最短 最后算出来ab的三分点 然后cd自然也就算出来了 不过知得一个注意的地方的是再求距离时需要加上一个0.000001吧 不加就错了 我也搞不懂了
*/

#include<iostream>#include<stdio.h>#include<algorithm>#include<math.h>#include<string.h>using namespace std;typedef struct Point{    double x;    double y;}point;point a,b,c,d;point bb,dd;double ab,cd;double p,q,r;double path(point m,point n){    return sqrt(0.0000001+pow(m.x - n.x,2.0) + pow(m.y - n.y,2.0));}double work(double t){    dd.x = d.x + (c.x - d.x)/cd * t * q;    dd.y = d.y + (c.y - d.y)/cd * t *q;    return t + path(bb,dd)/r;}double sanfen1(double t){    bb.x = a.x + (b.x - a.x)/ab* t* p;    bb.y = a.y + (b.y - a.y)/ab * t* p;    double l = 0;    double r = cd/q;    double mid ,mmid;    while((r-l)>0.0000001){        mid = l + (r-l)/3;        mmid = r - (r-l)/3;        if(work(mid)<work(mmid)){            r = mmid;        }else            l = mid;    }    return t + work(l);}int main(){    int ncase;    cin>>ncase;    while(ncase--){        cin>>a.x>>a.y>>b.x>>b.y;        cin>>c.x>>c.y>>d.x>>d.y;        cin>>p>>q>>r;        ab = path(a,b);        cd = path(c,d);        double l =0,r = ab/p;        double mid;        double mmid;        while((r-l)>0.000001){             mid = l + (r - l)/3;             mmid = r - (r - l)/3;            if(sanfen1(mid)<sanfen1(mmid))                r = mmid;            else                l = mid;        }        printf("%.2f\n",sanfen1(mid));    }    return 0;}