poj 3253(哈夫曼树priority_queue实现)

Fence Repair

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 21011 Accepted: 6697

Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needsN (1 ≤N ≤ 20,000) planks of wood, each having some integer lengthL_i (1 ≤L_i ≤ 50,000) units. He then purchases a single long board just long enough to saw into theN planks (i.e., whose length is the sum of the lengthsL_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 theN-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 theN 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 makeN-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

哈夫曼算法的核心思想就是，按照预先设想的排序准则s，构建一棵二叉树。使得在排序准则s中排在前面的尽可能在树的底层，排在后面的元素尽可能在树的高层。

而且 所有初始结点都在叶结点上。

由二叉树的性质：当一棵树是二叉树时(每个结点或者没有子结点，或者有2个子结点)，叶结点数=内结点数+1

因此当结点为n时，需要n-1次合并操作。

huffman树可以用priority_queue实现。需要重载<操作符。主要操作是push, top, pop操作。

提交记录：

1.Wrong Answer。居然是因为没有用long long 。题目中说好的是20000*50000,不知道为何会暴int。莫名奇妙。

2.AC

代码如下：