HDU 3400 Line belt 嵌套三分搜索

Line belt

Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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Status

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

嵌套三分

//嵌套三分#include<iostream>#include<stdio.h>#include<cstring>#include<vector>#include<math.h>#include<sstream>#include<set>#include<map>#include<iomanip>#include<algorithm>using namespace std;const double MIN=0.00001;struct point{    double x,y;};double dis(point a,point b){    return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2));}point a,b,c,d;double p,q,r;double solve_ab(point m){    point left=a,right=b;    point mid,midd;    double ans1=2,ans2=1;    while (fabs(ans1-ans2)>MIN)    {        mid.x=(left.x+right.x)/2;        mid.y=(left.y+right.y)/2;        midd.x=(mid.x+right.x)/2;        midd.y=(mid.y+right.y)/2;        ans1=dis(a,mid)/p+dis(mid,m)/r;        ans2=dis(a,midd)/p+dis(midd,m)/r;        if (ans1<ans2)            right=midd;        else            left=mid;    }    return ans1;}double solve_abcd(){    point left=c,right=d;    point mid,midd;    double ans1=2,ans2=1;    while (fabs(ans1-ans2)>MIN)    {        mid.x=(left.x+right.x)/2;        mid.y=(left.y+right.y)/2;        midd.x=(mid.x+right.x)/2;        midd.y=(mid.y+right.y)/2;        ans1=dis(d,mid)/q+solve_ab(mid);        ans2=dis(d,midd)/q+solve_ab(midd);        if (ans1<ans2)            right=midd;        else            left=mid;    }    return ans1;}int main(){    cout<<fixed<<setprecision(2);    int T;    cin>>T;    while (T--)    {        cin>>a.x>>a.y>>b.x>>b.y;        cin>>c.x>>c.y>>d.x>>d.y;        cin>>p>>q>>r;        cout<<solve_abcd()<<endl;    }}