SDAU 搜索专题 06 Line belt

1：问题描述
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

2：大致题意

给出ABCD的坐标，和在 AB，CD，其他线段，上的移动速度。要求出从A到D的最短时间。

3：思路

每次三分AB的时候，都要通过3分法得到CD上的最优解。
路线一定是 这样的。 从 A开始 到 AB 上的 o 点，再到 cd 上的 n 点，最后到达 D 。 其中Ao nD 都可以为 0 。
列出表达式 Ao / p + on /v +nD/q。

4：感想

要双向3分。每次三分AB的时候，都要通过3分法得到CD上的最优解。
好难啊啊啊啊。参考的别人的博客。

5：ac代码

#include<iostream>#include<string.h>#include<set>#include<stdio.h>#include<vector>#include<algorithm>#include<numeric>#include<math.h>#include<string.h>#include<sstream>#include<stdio.h>#include<string>#include<cstdlib>#include<algorithm>#include<iostream>#include<map>#include<queue>#include<iomanip>#include<cstdio>using namespace std; struct point{    double x,y;};int p,q,r,v;double dis(point a,point b){    return pow(pow(a.y-b.y,2)+pow(a.x-b.x,2),1.0/2);}double findy(point c,point d,point y){    point mid,midmid,left,right;    double t1,t2;    left = c;    right = d;    do    {        mid.x = (left.x+right.x)/2;        mid.y = (left.y+right.y)/2;        midmid.x = (right.x+mid.x)/2;        midmid.y = (right.y+mid.y)/2;        t1 = dis(d,mid)/q+dis(mid,y)/r;        t2 = dis(d,midmid)/q+dis(midmid,y)/r;        if(t1>t2)        left = mid;        else right = midmid;    }while(fabs(t1-t2)>0.000001);    return t1;}double find(point a,point b,point c,point d){    point mid,midmid,left,right;    double t1,t2;    left = a;    right = b;    do    {        mid.x = (left.x+right.x)/2;        mid.y = (left.y+right.y)/2;        midmid.x = (right.x+mid.x)/2;        midmid.y = (right.y+mid.y)/2;        t1 = dis(a,mid)/p+findy(c,d,mid);        t2 = dis(a,midmid)/p+findy(c,d,midmid);        if(t1>t2)left = mid;        else right = midmid;    }while(fabs(t1-t2)>0.000001);    return t1;}int main(){    int t;    point a,b,c,d;    scanf("%d",&t);    while(t--)    {        scanf("%lf%lf%lf%lf %lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&d.x,&d.y);        scanf("%d%d%d",&p,&q,&r);        printf("%.2f\n",find(a,b,c,d));    }    return 0;}