Line belt

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 9 Accepted Submission(s) : 5

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

Author

lxhgww&&momodi

题目大意：在平面上有两条直线：AB，CD，从A到D，如果在直线AB上行驶有速度P，在直线CD上行驶有速度Q,其它地方有速度R，求A到D的最短花费时间。

解题思路：

首先固定AB上一点，第一次我们设把这点设为mid1=B，然后三分去CD上找一点mid2，使A到D花的时间最少。然后固定CD上的mid2，用三分法去AB上找一点mid1,使A->mid1->mid2->D,花费时间最少，重复上述过程。

不知道错哪了...

#include<cstdio>#include<cmath>#define min(a,b) (a>b?b:a)#define esp 1e-8double p,q,r;struct point{    double x,y;}A,B,C,D;double get_dis(point M,point N){    return sqrt((N.x-M.x)*(N.x-M.x)+(N.y-M.y)*(N.y-M.y));}point get_mid(point M,point N){    point mid;    mid.x=(N.x+M.x)/2.0;    mid.y=(N.y+M.y)/2.0;    return mid;}double get_time(point M,point N){    return get_dis(A,M)/p+get_dis(M,N)/r+get_dis(N,D)/q;}double three2(point N){    point min=D,max=C,M2,M3;//二分点，三分点    double time2,time3;    while(get_dis(min,max)>esp)    {        M2=get_mid(min,max);time2=get_time(N,M2);        M3=get_mid(M2,max);time3=get_time(N,M3);        if(time2<time3)          max=M3;        else          min=M2;    }    return min(time2,time3);}using namespace std;int main(){    int cases;    scanf("%d",&cases);    while(cases--)    {        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);        point min=A,max=B,M2,M3;        double time2,time3;        while(get_dis(min,max)>esp)        {            M2=get_mid(min,max);            M3=get_mid(M2,max);            time2=three2(M2);            time3=three2(M3);            if(time2>time3)              min=M2;            else              max=M3;        }        printf("%.2lf\n",min(time2,time3));    }    return 0;}