Fence Repair

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the “kerf”, the extra length lost to sawdust when a sawcut is made; you should ignore it, too.
FJ sadly realizes that he doesn’t own a saw with which to cut the wood, so he mosies over to Farmer Don’s Farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn’t lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.
Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.

Input

Line 1: One integer N, the number of planks
Lines 2.. N+1: Each line contains a single integer describing the length of a needed plank
Output
Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts

Sample Input

3
8
5
8

Sample Output

34

分析

总共有n-1 次是固定的, 所以我们要让大的块所在的块尽量少切
这就利用了哈夫曼编码的原理, 先排序,取出来块最小的两个, 然后合并入优先队列(堆), 这样
最小得所在的块切的次数多, 大的次数少
特别注意 int 会溢出

具体实现如下


int main()
{

    priority_queue<LL,vector<LL>,greater<LL> > q;
    int N;
    cin>>N;
    while(N--)
    {
        LL a;
        scanf("%I64d",&a);
        q.push(a);
    }
    long long  sum = 0;//sum即是所求的费用
    while(q.size()!=1)
    {
         LL a  = q.top();
         q.pop();
         LL b = q.top();//取出最小的两块
         q.pop();
         sum += a+b;
         q.push(a+b);//将最小的两块合并堆
    }
    cout<<sum<<endl;
// cout<<q.top()<<endl;
     return 0;
}