uva10954Add All(贪心+优先队列)

最新推荐文章于 2019-11-09 10:13:42 发布

原创最新推荐文章于 2019-11-09 10:13:42 发布 · 822 阅读

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本文探讨了一种通过优先队列优化的贪心算法来解决最小成本加和问题。通过实例解析，阐述了如何在一系列整数中进行加法操作以达到最小总成本。详细介绍了输入格式、输出要求及代码实现，旨在提升算法理解和实践能力。

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题目链接：http://vjudge.net/contest/view.action?cid=50494#problem/D

Add All
Input: standard input
Output: standard output

Yup!! The problem name reflects your task; just add a set of numbers. But you may feel yourselves condescended, to write a C/C++ program just to add a set of numbers. Such a problem will simply question your erudition. So, let’s add some flavor of ingenuity to it.

Addition operation requires cost now, and the cost is the summation of those two to be added. So, to add 1 and 10, you need a cost of 11. If you want to add 1, 2 and 3. There are several ways –

1 + 2 = 3, cost = 3

3 + 3 = 6, cost = 6

Total = 9

1 + 3 = 4, cost = 4

2 + 4 = 6, cost = 6

Total = 10

2 + 3 = 5, cost = 5

1 + 5 = 6, cost = 6

Total = 11

I hope you have understood already your mission, to add a set of integers so that the cost is minimal.

Input

Each test case will start with a positive number, N (2 ≤ N ≤ 5000) followed by N positive integers (all are less than 100000). Input is terminated by a case where the value of N is zero. This case should not be processed.

Output

For each case print the minimum total cost of addition in a single line.

Sample Input Output for Sample Input

1 2 3

1 2 3 4

题目的意思是每次只能使两个数相加此次的和为这次操作的花费最后要使所有的数加起来求最小的花费，很明显是一道贪心的题目，但是每次加法操作产生的结果也要参与进来，因此我们需要用到优先队列

代码如下：

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
int main()
{
    int n;
    while(cin>>n,n){
        int a;
        priority_queue<long long, vector<long long>, greater<long long> >q;//从小到大
        //重载优先级,默认的是从大到小排列        
        for(int i=0;i<n;i++){
            cin>>a;
            q.push(a);
        }
        long long ans=0,x1,x2;
        while(!q.empty()){
           x1=q.top();
           q.pop();//注意只剩一个元素的时候
           if(q.empty())
             break;
           x2=q.top();
           q.pop();
           x1+=x2;
           ans+=x1;
           q.push(x1);
        }
        cout<<ans<<endl;
    }
    return 0;
}