HDU 3400 Line belt （三分嵌套）

题目链接

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2862    Accepted Submission(s): 1099

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

 

Author

lxhgww&&momodi

 

题意：

给出两条传送带的起点到末端的坐标，其中ab为p的速度，cd为q的速度 其他地方为r的速度

求a到d点的最短时间。

分析：

首先要看出来这是一个凹型的函数，

时间最短的路径必定是至多3条直线段构成的，一条在AB上，一条在CD上，一条架在两条线段之间。

所有利用两次三分，第一个三分ab段的一点，第二个三分知道ab一点后的cd段的接点。

刚开始没用do while错了两次，因为如果给的很接近的话，上来的t1没有赋值。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define LL __int64
const int maxn = 1e2 + 10;
const double eps = 1e-8;
using namespace std;
double p, q, r;
struct node
{
    double x, y;
}a, b, c, d;

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

double solve2(node t)
{
    double d1, d2;
    node le = c;
    node ri = d;
    node mid, midmid;
    do
    {
        mid.x = (le.x+ri.x)/2.0;
        mid.y = (le.y+ri.y)/2.0;
        midmid.x = (mid.x+ri.x)/2.0;
        midmid.y = (mid.y+ri.y)/2.0;
        d1 = dis(t, mid)/r + dis(mid, d)/q;
        d2 = dis(t, midmid)/r + dis(midmid, d)/q;
        if(d1 > d2)
        le = mid;
        else ri = midmid;
    }while(dis(le, ri)>=eps);
    return d1;
}

double solve1()
{
    double d1, d2;
    node le = a;
    node ri = b;
    node mid, midmid;
    do
    {
        mid.x = (le.x+ri.x)/2.0;
        mid.y = (le.y+ri.y)/2.0;
        midmid.x = (mid.x+ri.x)/2.0;
        midmid.y = (mid.y+ri.y)/2.0;
        d1 = dis(a, mid)/p + solve2(mid);
        d2 = dis(a, midmid)/p + solve2(midmid);
        if(d1 > d2)
        le = mid;
        else ri = midmid;
    }while(dis(le, ri)>=eps);
    return d1;
}

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);
        printf("%.2lf
", solve1());
    }
    return 0;
}