HDU 3400 line belt (three minutes and three minutes)

HDU 3400 line belt (three minutes and three minutes)

ACM

Question address:
HDU 3400 line belt

Question:
It is to give you two line segments AB and CD. A person runs P at the speed of AB, Q at the CD, and R at the speed of other places. Ask you the minimum time from A to D.

Analysis:
First, the point on the Three-Point AB, and then the point on the Three-Point CD.
Proof:
Set e to AB and F to CD.
The time spent on the online segment AB is:f = AE / pThe time taken by a person to finish Z and Y is:g = EF / r + FD / q.
F function is a monotonic increasing function, while G is obviously a function that decreases first and then increases progressively. The two functions are superimposed, and the resulting function should also be a function that decreases first and then increases progressively. Therefore, you can use the three-way solution.

Code:

/**  Author:      illuz <iilluzen[at]gmail.com>*  File:        3400.cpp*  Create Date: 2014-09-18 09:44:01*  Descripton:  triple*/#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const double EPS = 1e-8;struct Point {double x;double y;} a, b, c, d, e, f;int t;double p, q, r;double dis(Point p1, Point p2) {return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));}double calc(double alpha) {f.x = c.x + (d.x - c.x) * alpha;f.y = c.y + (d.y - c.y) * alpha;return dis(f, d) / q + dis(e, f) / r;}double inter_tri(double alpha) {e.x = a.x + (b.x - a.x) * alpha;e.y = a.y + (b.y - a.y) * alpha;double l = 0.0, r = 1.0, mid, mmid, cost;while (r - l > EPS) {mid = (l + r) / 2;mmid = (mid + r) / 2;cost = calc(mid);if (cost <= calc(mmid))r = mmid;elsel = mid;}return dis(a, e) / p + cost;}double solve() {double l = 0.0, r = 1.0, mid, mmid, ret;while (r - l > EPS) {mid = (l + r) / 2;mmid = (mid + r) / 2;ret = inter_tri(mid);if (ret <= inter_tri(mmid))r = mmid;elsel = mid;}return ret;}int main() {ios_base::sync_with_stdio(0);cin >> t;while (t--) {cin >> a.x >> a.y >> b.x >> b.y;cin >> c.x >> c.y >> d.x >> d.y;cin >> p >> q >> r;printf("%.2f\n", solve());}return 0;}

HDU 3400 line belt (three minutes and three minutes)