hdu 3400Line belt(三分法)

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2517    Accepted Submission(s): 961


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
 
題目大意:
給出兩條線段AB和CD,在AB上運動有一個速度,在CD上運動有一個速度,兩條線段之外有一個速度。現在從A點出發,到達D點。求所需最短時間。
解題思路:
三分法,找出AB上運動到D點的最短時間的點,再找出在CD上的點。那麼路線A到D之間經過這兩個點的時間就是最短的。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
#ifdef __int64
typedef __int64 LL;
#else
typedef long long LL;
#endif
#define eps 1e-8//注意精度問題
struct point
{
    double x,y;
};

double p,q,r;

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

double fund(point a,point c,point d)
{
    point mid;
    point midmid;
    point left=c;
    point right=d;
    double mid_value,midmid_value;
    int flag=0;
    while(dis(left,right)>=eps)
    {
        mid.x=(left.x+right.x)/2.0;
        mid.y=(left.y+right.y)/2.0;
        midmid.x=(mid.x+right.x)/2.0;
        midmid.y=(mid.y+right.y)/2.0;
        mid_value=dis(a,mid)/r+dis(mid,d)/q;
        midmid_value=dis(a,midmid)/r+dis(midmid,d)/q;
        if(mid_value<=midmid_value)
            right=midmid;
        else left=mid;
        flag=1;
    }
    if(flag)
        return mid_value;
    else
        return dis(a,d)/r;
}

double ans(point a,point b,point c,point d)
{
    point mid,midmid;
    point left=a;
    point right=b;
    double mid_value,midmid_value;
    int flag=0;
    while(dis(left,right)>=eps)
    {
        mid.x=(left.x+right.x)/2.0;
        mid.y=(left.y+right.y)/2.0;
        midmid.x=(mid.x+right.x)/2.0;
        midmid.y=(mid.y+right.y)/2.0;
        mid_value=dis(mid,a)/p+fund(mid,c,d);
        midmid_value=dis(midmid,a)/p+fund(midmid,c,d);
        if(mid_value<=midmid_value)
            right=midmid;
        else left=mid;
        flag=1;
    }
    if(flag)
        return mid_value;
    else
        return fund(a,c,d);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        point a,b,c,d;
        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);
        printf("%.2lf\n",ans(a,b,c,d));
    }
    return 0;
}


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