hdu 3400 Line belt 三分

题意：已知在线段AB上的速度是p，在CD上的速度是q，在其他地方的速度是r，求A到D的最短时间

思路：三分AB上一点，根据这点三分CD

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#include <vector>
#include <queue>
#include <math.h>
using namespace std;
#define LL long long
#define eps 1e-6
#define maxn 300
struct point
{
    double x,y;
    point()
    {

    }
    point(double xx,double yy)
    {
        x=xx;
        y=yy;
    }
} p[5];
double dist(point a,point b)
{
    return hypot(fabs(a.x-b.x),fabs(a.y-b.y));
}
double P,Q,R;
double solve(point c,point d,point x)
{
    point mid1,mid2,left,right;
    double t1=0,t2=1;
    left=c;
    right=d;
    while(fabs(t1-t2)>eps)
    {
        mid1=point((left.x+right.x)/2,(left.y+right.y)/2);
        mid2=point((right.x+mid1.x)/2,(right.y+mid1.y)/2);
        t1=dist(d,mid1)/Q+dist(x,mid1)/R;
        t2=dist(d,mid2)/Q+dist(x,mid2)/R;
        if(t1>t2) left=mid1;
        else right=mid2;
    }
    return t1;
}
int main()
{
    int T,i;
    scanf("%d",&T);
    while(T--)
    {
        for(i=0; i<4; i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        scanf("%lf%lf%lf",&P,&Q,&R);
        point mid1,mid2,left,right;
        double t1=0,t2=1;
        left=p[0];
        right=p[1];
        while(fabs(t1-t2)>eps)
        {
            mid1=point((left.x+right.x)/2,(left.y+right.y)/2);
            mid2=point((right.x+mid1.x)/2,(right.y+mid1.y)/2);
            t1=dist(p[0],mid1)/P+solve(p[2],p[3],mid1);
            t2=dist(p[0],mid2)/P+solve(p[2],p[3],mid2);
            if(t1>t2) left=mid1;
            else right=mid2;
        }
        printf("%.2f\n",t1);
    }
    return 0;
}