POJ 3253 切木板最小开销 贪心

POJ3253

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 71072		Accepted: 23372

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 Nplanks (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).

思路：

1. 一开始可能会错误地想到每次都砍掉一个最大的，然后只需要把所有的块从大到小排序就可以了，这样不对

举个例子：1 2 3 4 5

如果是从大到小排序：5 4 3 2 1，我每次都砍一个最大的，结果应该是：15 + 10 + 6 + 3 = 34

• 正确的思路是把砍木板看做拼合木板，每次都拼接两个最小的

还是：1 2 3 4 5，我每次都拼接两个最小的：

1+2 = 3 -> 3 3 4 5

3+3 = 6 -> 4 5 6（注意大小顺序）

4+5 = 9 -> 6 9

6+9 = 15-> 15

全过程的消耗为：3 + 6 + 9 + 15 = 33，要比每次都砍一个最大的少花1块

按照这个结果倒着想，正确的砍木块的顺序应该是：

第一刀：15砍成9+6

第二刀：9砍成4+5

第三刀：6砍成3+3

第四刀：3砍成2+1

难点：

• 贪心思路
• 每次都取最短的两个木板（注意最短的两个木板不一定是最后的两个木板）需要用到优先队列priority_queue

题解：

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