HDU 3400 Line belt (三分搜索)

Line belt

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

Total Submission(s): 2329 Accepted Submission(s): 880

[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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

[align=left]Output[/align]
The minimum time to travel from A to D, round to two decimals.

[align=left]Sample Input[/align]

1
0 0 0 100
100 0 100 100
2 2 1

[align=left]Sample Output[/align]

136.60

[align=left]Author[/align]
lxhgww&&momodi

[align=left]Source[/align]
HDOJ Monthly Contest – 2010.05.01

[align=left]Recommend[/align]
lcy

题意：

给你一个平面，平面上有两条线段AB和CD，某东西在AB上跑的速度是P，在CD上跑的速度是Q，在平面其他地方跑的速度是R，问从A点到D点的最快时间。

思路：

嵌套的三分搜索

1.首先明确需要在AB和CD上分别确定一个点作为转折点。

2.若已知AB，CD上的定点P1,P2,那么Time = AP1/P + P1P2/R + P2D/Q;

3.若我们能够先知道点P1(x1,y1),那么可以根据P2(x2,y2)求出最快的时间Time。

Time = AP1/P + sqrt((x1-x2)^2 + (y1-y2)^2)/R + sqrt((x2-D.x) + (y2-D.y))/Q;

/*************************************************************************
> File Name: hdu3400.cpp
> Author: BSlin
> Mail:
> Created Time: 2013年09月29日 星期日 22时07分59秒
************************************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <iterator>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#define MP make_pair
#define INF (1<<30)
#define PI acos(-1.0)
#define esp 1e-5
const int dx[4]={0,0,0,0};
using namespace std;
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)
#define LL __int64
#define LLS "%" "I" "6" "4" "d"
#else
#define LL long long
#define LLS "%" "l" "l" "d"
#endif

struct Point {
double x,y;
}A,B,C,D;

int P,Q,R;

Point Get_Point (Point A, Point B) {
Point Mid;
Mid.x = (A.x + B.x) / 2.0;
Mid.y = (A.y + B.y) / 2.0;
return Mid;
}

double Get_Len (Point A, Point B) {
return sqrt((A.x -B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}

double Get_Time (Point p) {
Point lift = C,right = D,mid,mmid;
double mid_time = 0,mmid_time = 1;
while(fabs(mid_time - mmid_time) > esp) {
mid = Get_Point(lift,right);
mmid = Get_Point(mid,right);
mid_time = Get_Len(p,mid) / R + Get_Len(mid,D) / Q;
mmid_time = Get_Len(p,mmid) / R + Get_Len(mmid,D) / Q;
if(mid_time - mmid_time >= esp) lift = mid;
else right = mmid;
}
return mid_time + Get_Len(A,p) / P;
}

int main(int argc, char** argv) {
//read;
int t;
scanf("%d",&t);
while(t--) {
scanf("%lf%lf%lf%lf%lf%lf%lf%lf%d%d%d",&A.x,&A.y,&B.x,&B.y,&C.x,&C.y,&D.x,&D.y,&P,&Q,&R);
Point lift = A,right = B,mid,mmid;
double mid_time = 0,mmid_time = 1;
while(fabs(mid_time - mmid_time) > esp) {
mid = Get_Point(lift,right);
mmid = Get_Point(mid,right);
mid_time = Get_Time(mid);
mmid_time = Get_Time(mmid);
if(mid_time - mmid_time >= esp) lift = mid;
else right = mmid;
}
printf("%.2f\n", mid_time);
}
return 0;
}