POJ 1862 & ZOJ 1543 Stripies（贪心 | 优先队列）

题目链接：

PKU:http://poj.org/problem?id=1862

ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=543

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：

就是有一些细胞条纹，他们会碰撞，碰撞后会变为一个新的细胞条纹，碰撞后的重量依照2*sqrt（m1*m2）计算，问最后的最小重量；

PS:

開始半天没读懂题意；

代码一例如以下：（贪心）

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
    int n;
    int a[117];
    while(~scanf("%d",&n))
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a,a+n);
        double tt = a[n-1];
        for(int i = n-2; i >= 0; i--)
        {
            tt = 2*sqrt(tt*a[i]);
        }
        printf("%.3lf\n",tt);
    }
    return 0;
}

代码二例如以下：（优先队列）

#include <cstdio>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
    int n;
    double a, b, c, tt;
    priority_queue<double,vector<double>,less<double> > Q;
    while(~scanf("%d",&n))
    {
        while(!Q.empty())
        {
            Q.pop();
        }
        for(int i = 0; i < n; i++)
        {
            scanf("%lf",&a);
            Q.push(a);
        }
        for(int i = 0; i < n-1; i++)
        {
            b = Q.top();
            Q.pop();
            c = Q.top();
            Q.pop();
            tt = 2*sqrt(b*c);
            Q.push(tt);
        }
        printf("%.3f\n",Q.top());
    }
    return 0;
}