Codeforces Round #327 (Div. 1) B. Chip 'n Dale Rescue Rangers 二分

题目链接：

题目

B. Chip 'n Dale Rescue Rangers
time limit per test：1 second
memory limit per test：256 megabytes

问题描述

A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signal of distress. In a few moments they were ready, and the dirigible of the rescue chipmunks hit the road.

We assume that the action takes place on a Cartesian plane. The headquarters of the rescuers is located at point (x1, y1), and the distress signal came from the point (x2, y2).

Due to Gadget's engineering talent, the rescuers' dirigible can instantly change its current velocity and direction of movement at any moment and as many times as needed. The only limitation is: the speed of the aircraft relative to the air can not exceed meters per second.

Of course, Gadget is a true rescuer and wants to reach the destination as soon as possible. The matter is complicated by the fact that the wind is blowing in the air and it affects the movement of the dirigible. According to the weather forecast, the wind will be defined by the vector (vx, vy) for the nearest t seconds, and then will change to (wx, wy). These vectors give both the direction and velocity of the wind. Formally, if a dirigible is located at the point (x, y), while its own velocity relative to the air is equal to zero and the wind (ux, uy) is blowing, then after seconds the new position of the dirigible will be .

Gadget is busy piloting the aircraft, so she asked Chip to calculate how long will it take them to reach the destination if they fly optimally. He coped with the task easily, but Dale is convinced that Chip has given the random value, aiming only not to lose the face in front of Gadget. Dale has asked you to find the right answer.

It is guaranteed that the speed of the wind at any moment of time is strictly less than the maximum possible speed of the airship relative to the air.

输入

The first line of the input contains four integers x1, y1, x2, y2 (|x1|,  |y1|,  |x2|,  |y2| ≤ 10 000) — the coordinates of the rescuers' headquarters and the point, where signal of the distress came from, respectively.

The second line contains two integers and t (0 < v, t ≤ 1000), which are denoting the maximum speed of the chipmunk dirigible relative to the air and the moment of time when the wind changes according to the weather forecast, respectively.

Next follow one per line two pairs of integer (vx, vy) and (wx, wy), describing the wind for the first t seconds and the wind that will blow at all the remaining time, respectively. It is guaranteed that and .

输出

Print a single real value — the minimum time the rescuers need to get to point (x2, y2). You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .

样例

input
0 0 5 5
3 2
-1 -1
-1 0

output
3.729935587093555327

题意

救援飞船在(x1,y1)点，遇难者在(x2,y2)点，在前t秒时间里刮一号风：(vx,vy)，在之后的时间里刮二号风(wx,wy)，你飞船的最大速度为v，v大于风的速度。问你飞船能够到达遇难者的最短的时间。

题解

首先，如果给我们的时间越多，我们就越能够接近飞船！(一个特殊情况就是把飞船的前进方向控制在直线上。）也就是说，时间在可达和不可达的问题上是单调的！，我们可以对时间进行二分！求出最小的可行解！

那么要怎么判断给定的时间t,飞船能不能到达遇难者的位置呢？
其实只要给定固定的时间，我们的最优行路线是固定的！
速度是可以叠加的，我们可以考虑在t秒的时间，只有风的作用下的情况，我们会从（x1,y1)到达（x1+tvx,y1+tvy)，记为(xm,ym)，接下来再考虑没有风的情况下用速度v是否能从(xm,ym)到达(xt,yt)就可以了，即判断v*t>=dis((xm,ym)->(xt,yt))是否成立;

那么有两段风怎么办？
我们可以考虑先判断一下t秒这个时间是否可以到达，如果可以，那我们二分的范围就是(0,t），且只要考虑第一个风。 如果不可以，就二分（t,INF),且两个风都要考虑（利用叠加的原理，其实和 上面一种情况没什么差。）。

代码

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

typedef __int64 LL;
int xs,ys,xt,yt,v,t,vx,vy,wx,wy;
const int INF=0x3f3f3f3f;
const int TIM=1000;
const double eps=1e-8;

int Scan(){ int x; scanf("%d",&x); return x; }

void input(){
	xs=Scan(); ys=Scan();
	xt=Scan(); yt=Scan();
	v=Scan();  t=Scan();
	vx=Scan(); vy=Scan();
	wx=Scan(); wy=Scan();
}

int main(){
	input();
	double xm=xs+vx*t,ym=ys+vy*t;
	double dis=sqrt((xt-xm)*(xt-xm)+(yt-ym)*(yt-ym));
	if(1.0*v*t>=dis+eps){
		double l=0,r=t;
		while(r-l>eps){
			double mid=l+(r-l)/2;
			double _xm=xs+vx*mid,_ym=ys+vy*mid;
			double _dis=sqrt((xt-_xm)*(xt-_xm)+(yt-_ym)*(yt-_ym));
			if(1.0*v*mid>=_dis+eps) r=mid;
			else l=mid;
		} 
		printf("%.18lf\n",r);
	}else{
		double l=0,r=INF;
		while(r-l>eps){
			double mid=l+(r-l)/2;
			double _xm=xm+wx*mid,_ym=ym+wy*mid;
			double _dis=sqrt((xt-_xm)*(xt-_xm)+(yt-_ym)*(yt-_ym));
			if(1.0*v*(mid+t)>=_dis+eps) r=mid;
			else l=mid;
		}
		printf("%.18lf\n",r+t);
	}
	return 0;
}