算法竞赛宝典分治算法交叉的梯子(几何公式的推导+二分)

最新推荐文章于 2021-04-27 21:12:41 发布

Jackie035

最新推荐文章于 2021-04-27 21:12:41 发布

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本文通过二分查找法解决了一类涉及梯子交叉高度的几何问题。具体地，给出了两个不同长度梯子在街道两侧相交时的具体计算方法，以求解街道宽度。通过精确的数学推导和边界条件处理，提供了高效的算法实现。

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Crossed ladders

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5209		Accepted: 1980

Description

A narrow street is lined with tall buildings. An x foot long ladder is rested at the base of the building on the right side of the street and leans on the building on the left side. A y foot long ladder is rested at the base of the building on the left side of the street and leans on the building on the right side. The point where the two ladders cross is exactly c feet from the ground. How wide is the street?

Input

Each line of input contains three positive floating point numbers giving the values of x, y, and c.

Output

For each line of input, output one line with a floating point number giving the width of the street in feet, with three decimal digits in the fraction.

Sample Input

30 40 10
12.619429 8.163332 3
10 10 3
10 10 1

Sample Output

Source

The UofA Local 2000.10.14

//理解二分边界处理+几何推导，这个就不难了，重点还是在边界

总结：做了几道类似的之后，二分对精度的处理都是，循环条件用精度划分，在二分时，直接left=mid，这样去处理

#include<iostream>
#include<algorithm>
#include<bits/stdc++.h>

using namespace std;
double x,y,c;
bool f(double w)
{
	double a,b;
	a=sqrt(x*x-w*w);
	b=sqrt(y*y-w*w);
	if(a*b>=c*(a+b))
		return 0;
	else
		return 1;
}
int main()
{
	while(cin>>x>>y>>c)
	{
		double left=0,right=x>y?y:x,mid;       //意思是以最短的楼梯作为街宽的最大然后进行二分
		while(right-left>1e-4)
		{
			mid=left+(right-left)/2.0;
			if(f(mid)==0)
				left=mid;
			else
				right=mid;	
		}
		printf("%.3lf\n",left);
	}
	return 0;
}