HDU 3400 Line belt 题解 三分镶嵌

Line belt

传送门

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

1
0 0 0 100
100 0 100 100
2 2 1

Output #1

136.60

题目翻译

题目描述

平面中两条线段 $AB$，$CD$，要求从 $A$ 点走到 $D$ 点，则先在 $AB$ 上走一段，然后在两线段之间走一段，再上 $CD$ 上走一段，三个地方的速度不同，求最短时间。

输入格式

第一行是案例编号 $T$。
对于每种情况，有三行。
第一行，四个整数，$A$ 和 $B$ 的坐标：$A_x,A_y,B_x,B_y$。
第二行，四个整数，$C$ 和 $D$ 的坐标：$C_x,C_y,D_x{D}_y$。
第三行，三个整数，$P,Q,R$。

输出格式

从 $A$ 运动到 $D$ 的最短时间，四舍五入到两位小数。

解题思路

使用三分法。因为从一点到另一条直线上的点的距离的变化不是单调的，而是先减后增，所以需要三分中镶嵌三分。
先三分枚举 $AB$ 上的点 $e$，或者时间，对应于点 $e$ 或者这个时间，去三分 $CD$ 段，找到对应 $AB$ 上当前点 $e$ 对应的使整体花费时间最小的 $BC$ 上的点 $f$，然后在 $AB$ 上移动一下，$BC$ 上搜一遍，继续移动，直到找到那个使整体花费时间最小的点，就是答案。

AC Code

// C++ includes used for precompiling -*- C++ -*-

// Copyright (C) 2003-2013 Free Software Foundation, Inc.
//
// This file is part of the GNU ISO C++ Library.  This library is free
// software; you can redistribute it and/or modify it under the
// terms of the GNU General Public License as published by the
// Free Software Foundation; either version 3, or (at your option)
// any later version.

// This library is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// Under Section 7 of GPL version 3, you are granted additional
// permissions described in the GCC Runtime Library Exception, version
// 3.1, as published by the Free Software Foundation.

// You should have received a copy of the GNU General Public License and
// a copy of the GCC Runtime Library Exception along with this program;
// see the files COPYING3 and COPYING.RUNTIME respectively.  If not, see
// .

/** @file stdc++.h
 *  This is an implementation file for a precompiled header.
 */

// 17.4.1.2 Headers

// C
#ifndef _GLIBCXX_NO_ASSERT
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#endif
using namespace std;
#define int long long
const double eps = 1e-9;
struct project {
	double x;
	double y;
} a, b, c, d, e, f;
double p, q, r;
double disab, discd;
inline double dis(project A, project B) {
	return sqrt(eps + (A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
inline double calc(double t) {
	f.x = d.x + (c.x - d.x) / discd * t * q;
	f.y = d.y + (c.y - d.y) / discd * t * q;
	return t + dis(e, f) / r;
}
inline double findf(double t) {
	e.x = a.x + (b.x - a.x) / disab * t * p;
	e.y = a.y + (b.y - a.y) / disab * t * p;
	double l = 0, r = discd / q, lmid, rmid;
	while (l + eps < r) {
		lmid = l + (r - l) / 3;
		rmid = r - (r - l) / 3;
		if (calc(lmid) < calc(rmid)) {
			r = rmid;
		} else {
			l = lmid;
		}
	}
	return t + calc(l);
}
inline double finde() {
	double l = 0.0, r = disab / p, lmid, rmid;
	while (l + eps < r) {
		lmid = l + (r - l) / 3;
		rmid = r - (r - l) / 3;
		if (findf(lmid) < findf(rmid)) {
			r = rmid;
		} else {
			l = lmid;
		}
	}
	return lmid;
}
inline void solve() {
	cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y >> d.x >> d.y >> p >> q >> r;
	disab = dis(a, b);
	discd = dis(c, d);
	printf("%.2f\n", findf(finde()));
}
inline void work() {
	int T;
	cin >> T;
	while (T--) {
		solve();
	}
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	work();
	return 0;
}

你以为写完了？对，确实。