贪心 + 数学 (POJ 1862)

Stripies

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 17148 Accepted: 7764

Description

Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, but the scientists had to invent an English name to apply for an international patent). The stripies are transparent amorphous amebiform creatures that live in flat colonies in a jelly-like nutrient medium. Most of the time the stripies are moving. When two of them collide a new stripie appears instead of them. Long observations made by our scientists enabled them to establish that the weight of the new stripie isn't equal to the sum of weights of two disappeared stripies that collided; nevertheless, they soon learned that when two stripies of weights m1 and m2 collide the weight of resulting stripie equals to 2*sqrt(m1*m2). Our chemical biologists are very anxious to know to what limits can decrease the total weight of a given colony of stripies. 
You are to write a program that will help them to answer this question. You may assume that 3 or more stipies never collide together. 

Input

The first line of the input contains one integer N (1 <= N <= 100) - the number of stripies in a colony. Each of next N lines contains one integer ranging from 1 to 10000 - the weight of the corresponding stripie.

Output

The output must contain one line with the minimal possible total weight of colony with the accuracy of three decimal digits after the point.

Sample Input

3723050

Sample Output

120.000

Source

Northeastern Europe 2001, Northern Subregion

有n个球，每两个球（假设质量为m1和m2）碰撞，质量变为2 * sqrt(m1 * m2)。三个球无法一起碰撞。问n个球不断碰撞后最小质量为多少？

贪心策略：每次选择质量最大的两个球。证明：假设a>=b>=c，那么2 * sqrt(2 * c * sqrt(a * b)) <= 2 * sqrt(2 * a * sqrt(b * c))。

代码实现使用优先队列。

// test.cpp : Defines the entry point for the console application.//// #include "stdafx.h"#include <cstdio>#include <iostream>#include <queue>#include <cmath>using namespace std;int n;int main(){while (scanf("%d", &n) == 1){priority_queue<double> q;double ans;for (int i = 1; i <= n; i++){scanf("%lf", &ans);q.push(ans);}while (q.size() > 1){double a = q.top();q.pop();double b = q.top();q.pop();double temp = 2 * sqrt(a * b);q.push(temp);}printf("%.3f\n", q.top());}    return 0;}