POJ 1862 Stripies【哈夫曼/贪心/优先队列】

Stripies

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 18198		Accepted: 8175

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

 

 

【题意】：给出n个物体，分别给出每个的质量，并且两个物体(假设质量分别为m1，m2)相撞的时候变成一个物体，质量为2*sqrt（m1*m2），并且只会出现两个两个物品碰撞的情况，问最终能得到的物体的最小质量是多少。

【分析】：类似于哈弗曼树的方法，每次选举两个当前最大的数值进行运算，直到剩下一个物品。使用优先队列处理比较方便。

【代码】：

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

#define LL long long
using namespace std;
const int N = +;
int n;
double L;
double a,b;
priority_queue<double> q;
int main()
{
    cin>>n;

        for(int i=;i<n;i++)
        {
            cin>>L;
            q.push(L);
        }
        while(q.size()>)
        {
            a=q.top();q.pop();

            b=q.top();q.pop();

            q.push(*sqrt(a*b));
        }
    printf("%.3f\n",q.top());

    return ;
}