poj 1862 Stripies（优先队列）

Stripies

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12528		Accepted: 5932

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

题意：给出n个数 每次取出两个数 m1 m2
     通过 2*sqrt(m1*m2) 进行合并
     问最后能得到的最大值为多少

思路：

这题其实可以直接排序的 结果考虑不周又想不起来优先队列怎么做 
贴个优先队列的代码当复习吧

排序的话 因为每次选出最大的两个数 合并的结果一定还是最大 所以可以按排序后的数组来合并

#include<iostream>
#include<vector>
#include<queue>
#include<cmath>
using namespace std;
int main()
{
  int N,M;
  double a,b,c,tmp;
  priority_queue<double> Q;
  scanf("%d",&N);
  M=N;
  while(N--)
  {
    scanf("%lf",&a);
    Q.push(a);
  }
  M--;
  while(M--)
  {
    b=Q.top();
    Q.pop();
    c=Q.top();
    Q.pop();
    tmp=2*sqrt(b*c);
    Q.push(tmp);
  }
  printf("%.3f\n",Q.top());
  //system("pause");
}