题目来源：http://poj.org/problem?id=1862
Stripies
| Time Limit: 1000MS |
Memory Limit: 30000K |
|
| Total Submissions: 19699 |
Accepted: 8800 |
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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思路
【题意】
有N个动物，每个有一个重量m, 重量为m1,m2的两个动物合成重量为2*sqrt(m1*m2)的动物，问所有N个动物全部合成之后的最小重量是多少
【思路】
最大优先队列，总是最大的两两合成合成出来的总重量最小。
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代码
// 最大优先队列
#include<cstdio>
#include<queue>
#include<cmath>
using namespace std;
const int NMAX = 105;
int main()
{
#ifndef ONLINE_JUDGE
freopen("1862.txt","r",stdin);
#endif
int n,i;
double a,b;
scanf("%d", &n);
priority_queue<double> q;
for (i=0; i<n; i++)
{
scanf("%lf", &a);
q.push(a);
}
while (q.size()>1)
{
a = q.top();
q.pop();
b = q.top();
q.pop();
a = 2*sqrt(a*b);
q.push(a);
}
printf("%.3lf",q.top());
return 0;
}