# POJ 3253 Fence Repair （哈夫曼）

Fence Repair

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 45098 Accepted: 14700

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 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

### 题意

FJ需要修补牧场的围栏，他需要 N 块长度为 Li 的木头（N planks of woods）。

### AC 代码

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
typedef __int64 LL;

int main()
{
int n;
while(~scanf("%d",&n))
{
priority_queue<int,vector<int>,greater<int> >sk;    //优先队列 小顶堆
int num;
LL ans=0;
for(int i=0; i<n; i++)
{
scanf("%d",&num);
sk.push(num);
}
if(sk.size()==1)    //如果只有1根木头
ans=sk.top();
while(sk.size()>1)
{
int a=sk.top();
sk.pop();
a+=sk.top();    //选取最小的两个求和
ans+=a;
sk.pop();
sk.push(a); //加入队列
}
printf("%I64d\n",ans);
}
return 0;
}