本文主要是介绍poj 3253 Fence Repair，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目链接：点击打开链接
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
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank
Output
Sample Input
3
8
5
8
Sample Output
34
Hint
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).
题目大意：有一个长木板，要分成n份，每次据开的时候要收取在据开之前的长度的钱 问怎么据才能花费最小，最小是多少
基本思路：主要是哈夫曼树的思想．，先取两个最短的，相加，将和加入花费中，再将和放入队列中，这里用优先队列，从小到大排序的．
虽然优先队列不是很会，但最简单的还是可以的．我认为用set函数也可以，当然用mulset(),可能会有重复的数
下面是代码．
<span style="font-size:24px;">#include <iostream>
#include<cstdio>
#include<queue>
#include<cstdlib>
#include<cstring>using namespace std;int main()
{priority_queue< int,vector<int >,greater<int> >q;//优先队列从小到大，great改为less表示从大到校int n;int x;cin>>n;for(int i=0;i<n;i++){cin>>x;q.push(x);}long long sum=0;while(q.size()>1){int a=q.top();q.pop();int b=q.top();q.pop();int c=a+b;sum+=c;q.push(c);}cout<<sum<<endl;return 0;
}
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