Add All - UVa 10954 优先队列

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 of11. 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

3 1 2 3 4 1 2 3 4 0

9 19

题意：每次将两个数加起来，和即为花费，问将所有数两两相加的最小花费。

思路：类似于哈弗曼树吧，每次找两个最小的数相加，然后将结果放入队列中，重复进行操作，直到只剩一个数位为止。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
struct node
{
    ll num;
    bool operator<(const node A)const
    {
        return A.num<num;
    }
};
priority_queue<node> qu;
ll num,ans;
int n;
node A,B,C;
int main()
{
    int i,j,k;
    while(~scanf("%d",&n) && n>0)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&num);
            A.num=num;
            qu.push(A);
        }
        ans=0;
        while(true)
        {
            A=qu.top();
            qu.pop();
            if(qu.empty())
              break;
            B=qu.top();
            qu.pop();
            num=A.num+B.num;
            ans+=num;
            C.num=num;
            qu.push(C);
        }
        printf("%lld\n",ans);
    }
}