B

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
Hint
He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.
The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).

一开始思路错了，一个是不熟悉，再就是跟着他给的例子走了，根本没能先从这个题的问法上进行思考。一开始想的是每次先把最大的木板去掉，这样不就是最小了，可是没有证明，只是猜测。看了别人的博客，才恍然大悟这是最小生成树，之后模拟就好了。可是，如果直接模拟的话，会超时，借助优先队列的优点，插入一个点后自动快速进行排序，保持序列的有序性。

#include<stdio.h>#include<queue>#define N  20010typedef  long long ll;using namespace std;int a[N];ll sum=0,temp1,temp2,temp,flag=0,ans=0;int main(){   priority_queue<ll,vector<ll>,greater<ll> > Queue;   int n,i;   scanf("%d",&n);   if(n==1)   {       scanf("%d",&a[0]);       printf("%lld\n",a[0]);   }   else   {        for(i=0;i<n;i++)        {            scanf("%d",&a[i]);            Queue.push(a[i]);        }      /*        while(!Queue.empty())        {            printf("%lld\n",Queue.top());            Queue.pop();        }       */    while(Queue.size()!=1)    {        //printf("%d*\n",Queue.top());       temp1=Queue.top();       Queue.pop();       temp2=Queue.top();       Queue.pop();       ans=ans+temp1+temp2;       Queue.push(temp1+temp2);    }        printf("%lld\n",ans);   }   return 0;}