Stripies (POJ 1862)

Time Limit: 1000MS Memory Limit: 30000K

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.

题目大意：

给定N个整数（1到100），从中选出2个数m1，m2进行合并，合并后的数字m = 2*sqrt(m1*m2)；重复执行合并操作，直到只剩下一个数字为止。

求最终得到数字的最小可能值

解题思路：

使用STL优先队列(priority_queue)，默认是大顶堆。每次选出最大的前2个数进行合并，将得到的数重新入队，依次类推。

C++代码：

#include <cstdio>
#include <cmath>
#include <queue>
std::priority_queue <double> pqueue;
int main()
{
  int tt,num,k;
  double a,b;
  while (scanf("%d",&tt) != EOF)
  {
    k = tt - 1;
    while (tt--)
    {
      scanf("%d",&num);
      pqueue.push(num);
    }
    while (k--)
    {
      a = pqueue.top();
      pqueue.pop();
      b = pqueue.top();
      pqueue.pop();
      pqueue.push(2 * (double)sqrt((double)(a * b)));
    }
    printf("%.3f\n",pqueue.top());
    pqueue.pop();
  }

  return 0;
}

 

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