HDU 3700 Line belt

Line belt

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3669 Accepted Submission(s): 1407



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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string.h>
#include <algorithm>

using namespace std;

const double eps = 1e-6;
double p,q,rr;
struct node{
double x,y;
};

double dis(node a,node b){
return sqrt( (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) );
}

//a - midd - mid - b
double tritri(node a,node b,node m)
{
node l,r;
l = a , r = b;
node mid,midd;

double ans1,ans2;

while(1){
mid.x = (l.x+r.x)/2.0;
mid.y = (l.y+r.y)/2.0;

midd.x = (l.x+mid.x)/2.0;
midd.y = (l.y+mid.y)/2.0;

ans1 = dis(mid,a)/q + dis(mid,m)/rr;
ans2 = dis(midd,a)/q + dis(midd,m)/rr;
if(ans1<ans2)
l = midd;
else
r = mid;

if( fabs(ans1-ans2)< eps)
return ans1;
}

}

//c - midd - mid - 0 - d
double tri (node a,node b,node c,node d){
node l,r;
l = c , r = d;
node mid,midd;

double ans1,ans2;

while(1){
mid.x = (l.x+r.x)/2.0;
mid.y = (l.y+r.y)/2.0;

midd.x = (l.x+mid.x)/2.0;
midd.y = (l.y+mid.y)/2.0;

ans1 = dis(mid,d)/q + tritri(a,b,mid);
ans2 = dis(midd,d)/q + tritri(a,b,midd);
if(ans1<ans2)
l = midd;
else
r = mid;

if( fabs(ans1-ans2)< eps)
return ans1;
}
}

int main()
{
int i,j,k,t,m,n;
node a,b,c,d;

cin>>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("%lf%lf%lf",&p,&q,&rr);
double ans = tri(a,b,c,d);
printf("%.2lf",ans);
cout<<endl;
}
return 0;
}