Hdu 3400 Line belt （三分）

题目链接

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2927    Accepted Submission(s): 1119

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
  

题意：两条线段AB和CD。起点在A，终点在D。在线段AB内的速度为p，在线段CD内的速度为q，在其它区域的速度为R。求从起点到终点的最短时间。

题解：先三分在AB线段的终点，在三分CD线段的起点。

代码如下：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<string>
#include<stack>
#include<math.h>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#define nn 110000
#define inff 0x7fffffff
#define eps 1e-8
#define mod 1000000007
typedef long long LL;
const LL inf64=LL(inff)*inff;
using namespace std;
struct point
{
    double x,y;
    point(double x=0,double y=0):x(x),y(y){}
};
point operator + (point A,point B)
{
    return point(A.x+B.x,A.y+B.y);
}
point operator/(point A,double B)
{
    return point(A.x/B,A.y/B);
}
point operator - (point a,point b)
{
    return point(a.x-b.x,a.y-b.y);
}
double dis(point A,point B)
{
    return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}
point A,B,C,D;
double p,q,r;
double check(point X,point Y)
{
    return dis(A,X)/p+dis(X,Y)/r+dis(Y,D)/q;
}
double solve(point X)
{
    point l=C,r=D;
    point mid,mmid;
    double ix,fc;
    while(dis(l,r)>eps)
    {
        mid=l+(r-l)/3.0;
        mmid=r-(r-l)/3.0;
        ix=check(X,mid);
        fc=check(X,mmid);
        if(ix>fc)
        {
            l=mid;
        }
        else
            r=mmid;
    }
    return check(X,l);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        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 l=A,r=B;
        point mid,mmid;
        double ix,fc;
        while(dis(l,r)>eps)
        {
            mid=l+(r-l)/3.0;
            mmid=r-(r-l)/3.0;
            ix=solve(mid),fc=solve(mmid);
            if(ix>fc)
            {
                l=mid;
            }
            else
                r=mmid;
        }
        printf("%.2lf\n",solve(l));
    }
    return 0;
}