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\n", solve(le));
    }
    return 0;
}