Fence Repair （优先队列）

Fence Repair

Time Limit: 2000MS
Memory Limit: 65536KTotal Submissions: 51161
Accepted: 16798

Description

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 L_i (1 ≤ L_i ≤ 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 L_i). 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

3858

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).

Source

USACO 2006 November Gold

题意： 比如8 5 8 先挑两个合成一个花费代价为两个之和，合成完了再放进去，重复这一步，直到只剩下一个为止，求花费的总代价的最小值；

这题当然是先选两个最下的合成后，再选最小的两个，直到只剩下一个为止；我是用了个优先队列 ，先全部加入，在取出最下的两个求和，在加去队列，s用来累加代价的；

#include<iostream>
#include<queue>
using namespace std;
int n,a;
priority_queue<int ,vector<int>, greater<int> >q;
void solve()
{
long long int s=0;
int s1,s2;
while(q.size()!=1)
{
s1=q.top();
q.pop();
s2=q.top();
q.pop();
int l=s1+s2;
s=s+l;
q.push(l);
}
cout<<s<<endl;
q.pop();
}
int main()
{


        while(cin>>n)
        {


            if(n==1)
            {
                cin>>a;
                cout<<a<<endl;
                continue;
            }
            while(n--)
            {
                cin>>a;
                q.push(a);
            }
            solve();
        }
return 0;
}

人一我百！人十我万！永不放弃~~~怀着自信的心，去追逐梦想。