HDU-3400Line belt-三分再三分-求距离中要加esp

传送门：Line belt

参考：http://blog.csdn.net/hcbbt/article/details/39375763

题意：在一个平面途中，有一条路ab，还有一条路cd；假设在ab，cd和其他地方的运动速度不同；

　　求从a到d的最短时间；

思路：三分在ab上的点，在三分cd上的点，找到对应最小的运动时间；

#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;
const double esp = 1e-6;

struct node {
    double x,y;
};
node a,b,c,d,b2,d2;
double ab,cd;
double p,q,r;
double dis(node a, node b)
{
    return sqrt(esp+(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));//就是这里要加入esp
}
double get(double t)
{
    d2.x= d.x+(c.x-d.x)/cd*t*q;
    d2.y= d.y+(c.y-d.y)/cd*t*q;
    return t + dis(b2,d2)/r;
}
double solve(double s)
{
    b2.x = a.x+(b.x-a.x)/ab*s*p;
    b2.y = a.y+(b.y-a.y)/ab*s*p;
    double le=0.0,ri=cd/q;
    while(le+esp<ri)
    {
        double m1=le+(ri-le)/3;
        double m2=ri-(ri-le)/3;
        if(get(m1)<get(m2))
        {
            ri=m2;
        }
        else le = m1;
    } 
    return s+get(le);
}
int main(){
    //freopen("in","r",stdin);
    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);
        ab = dis(a, b);
        cd = dis(c, d);
        double le=0.0, ri=ab/p;
        while(le+esp<ri)
        {
            double m1=le+(ri-le)/3;
            double m2=ri-(ri-le)/3;
            if(solve(m1)<solve(m2))
            {
                ri=m2;
            }
            else le = m1;
        } 
        printf("%.2lf
",solve(le));
    }
    return 0;
}