HDU 1969 && POJ 3122

原创于 2019-05-09 22:15:56 发布 · 212 阅读

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#HDU #POJ #二分查找

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本文介绍了一种使用二分查找算法解决蛋糕公平分配问题的方法。在庆祝生日时，如何将多个大小不一的圆形蛋糕切成相同体积的块，以确保每个人（包括主人）都能得到等量的一块？通过定义检查函数和实施二分查找，文章详细解释了如何找到最大可能的蛋糕体积，使得所有人都能获得同样大小的蛋糕块。

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Problem Description

My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

Input

One line with a positive integer: the number of test cases. Then for each test case:
---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.
---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.

Output

For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).

Sample Input

3 3 3 4 3 3 1 24 5 10 5 1 4 2 3 4 5 6 5 4 2

Sample Output

25.1327 3.1416 50.2655

思路：这是最近在看的基本算法，二分查找题目。首先一个check() 函数检验每次传进来的体积 volume 是否符合要求，标准就是看每一块Pie分成volume大小的可以分几块，所有Pie分块数求和看是否大于人数，因为Pie不能拆开，所以用地板除 floor(), （定义在cmath中）。然后是二分查找算法，从left = 0， right = sum(volume) 开始，逐步二分，最终找到最大的Volume。

这里有几个注意点，否则会WA：

PI = acos(-1.0); PI精度不够会出错（尝试了一下，PI貌似至少要10位）
每次二分时，left和right 要分别加减0.000001，也是精度问题。（OpenJudge上有道题Pie类似的，但精确不同）

下面是AC代码：

#include<iostream>
#include<vector> 
#include<cmath>
#include<algorithm>
#include<stdio.h>
using namespace std;

int N,F;
const double PI = acos(-1.0);
//#define PI 3.141592653589

double cal(int r)      //calculate the volume of a pies
{
	return PI * r * r;
}

bool check(double volume, vector<int> &p)       //check if satisfy the question
{
	int count = 0;
	for(int i = 0; i < N; ++i)
	{
		count += floor(cal(p[i]) / volume);
		
		if(count >= F)
			return true;
	}
	
	return false;
} 
 
double Binary_Search(double left, double right, vector<int> &p)     //search the biggest volume
{
	double mid = 0;
	
	while(left  <= right)
	{
		mid = left + (right-left) / 2;
		
		if(check(mid, p))
			left = mid + 0.000001;
		else
			right = mid - 0.000001;
	}
	
	return left-0.000001;        //最后一次check(true)多加的要减去 
}

int main(void){
	int T;
	cin >> T;
	while(T--){
		vector<int> p;
		
		cin >> N >> F;
		F++;                           //还有自己本人+1 
		for(int i = 0; i < N; ++i)
		{
			int temp;
			cin >> temp;
			p.push_back(temp);
		}
	
		sort(p.begin(), p.end());
	
		double right = 0;
		for(int i = 0; i < N;++i)
		{
			right += cal(p[i]);
		}
		
		printf("%.4lf\n", Binary_Search(0,right, p));
	}

	
	return 0;
}