【POJ】1862：Stripies【贪心】【优先队列】

Stripies

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 20456		Accepted: 9098

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

3
72
30
50

Sample Output

120.000

Source

Northeastern Europe 2001, Northern Subregion

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Solution

题目是求按$2*\sqrt{m1*m2}$两两合并能得到的最小值

假设有$a，b，c $且结果是$r$ 则 $r = 2*\sqrt{2*\sqrt{a*b}*c}$ 则$\frac{r^2}{8}=\sqrt{a*b*c*c}$若要 $r$ 最小 则 $c$ 一定是$a,b,c$中最小的 所以就是不断地取两个大数相乘

每次贪心取最大的两个元素合并即可....用优先队列实现吧....

（话说为什么poj上面必须选C++才过得了啊！！！还找了白天错QAQ

Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#define DB double
using namespace std;

int n;
DB a;
priority_queue < DB > q;

int main() {
    scanf("%d", &n);
    for(int i = ; i <= n; i ++)    scanf("%lf", &a), q.push(a);
    while(q.size() > ) {
        DB x = q.top(); q.pop();
        DB y = q.top(); q.pop();
        DB now =  * sqrt(x * y);
        q.push(now);
    }
    printf("%0.3lf\n", q.top());
    return ;
}