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挑道UVA的题题给大家玩玩, UVA 10003 - Cutting Sticks

LeeMaRS 2003-10-17 11:22:21
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Cutting Sticks

You have to cut a wood stick into pieces. The most affordable company, The Analog Cutting Machinery, Inc. (ACM), charges money according to the length of the stick being cut. Their procedure of work requires that they only make one cut at a time.
It is easy to notice that different selections in the order of cutting can led to different prices. For example, consider a stick of length 10 meters that has to be cut at 2, 4 and 7 meters from one end. There are several choices. One can be cutting first at 2, then at 4, then at 7. This leads to a price of 10 + 8 + 6 = 24 because the first stick was of 10 meters, the resulting of 8 and the last one of 6. Another choice could be cutting at 4, then at 2, then at 7. This would lead to a price of 10 + 4 + 6 = 20, which is a better price.

Your boss trusts your computer abilities to find out the minimum cost for cutting a given stick.


Input
The input will consist of several input cases. The first line of each test case will contain a positive number l that represents the length of the stick to be cut. You can assume l < 1000. The next line will contain the number n (n < 50) of cuts to be made.
The next line consists of n positive numbers ci ( 0 < ci < l) representing the places where the cuts have to be done, given in strictly increasing order.

An input case with l = 0 will represent the end of the input.


Output
You have to print the cost of the optimal solution of the cutting problem, that is the minimum cost of cutting the given stick. Format the output as shown below.

Sample Input
100
3
25 50 75
10
4
4 5 7 8
0


Sample Output
The minimum cutting is 200.
The minimum cutting is 22.

--------------------------------------------------------------------------------

Miguel Revilla
2000-08-21

大家慢慢考虑一下吧:)
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levinjoe 2003-10-20
是啊,算法方面国内就他写的还可以!
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levinjoe 2003-10-19
呵呵,是自己当初看书不认真,今天把<<计算机算法设计与分析》书中讲的四边形不等式又看了一遍!
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LeeMaRS 2003-10-19
嗯。若一个DP是对角线形的,且常数函数w满足四边形不等式,则基本上都能进行优化,降低复杂度。

《计算机算法设计与分析》是王晓东的书吧?
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galois_godel 2003-10-18
就是DP,不过改进了决策转移的时间
原来求c[i,j]=min(c[i,k]+c[k,j])+len[i,j];
要把 k 从 i 到 j 都扫描一边,效率为O(n)
现在 我们从题目性质可以证明
这个k是在 s[i][j-1] 到 s[i+1][j] 这个范围里中
(其中 ,s[i][j]是决策变量)
所以,转移的效率是常数。
再加上对O(n^2)个c[i,j]都附值一次,
总效率即为O(n^2)
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levinjoe 2003-10-18
to galois_godel()
能谈谈0(n^2)方法吗?我想了一下,没什么思路的,当初提交时是用动态规划做的!
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LeeMaRS 2003-10-17
:) 程序可以自行去UVA提交测试, 在这里就不要帖出来了, 偶希望的是大家描述自己的算法和思路.
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levinjoe 2003-10-17
呵呵,两年前提交通过了,用动态规划应该没问题!

今天有个晚会,下午要排练,没时间编了!留给楼下的兄弟吧!
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wengzhong 2003-10-17
动态规划?
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LeeMaRS 2003-10-17
嗯, 前段时间ZJU倒了, 就到UVA去摸了一把.
没想到这么快就被弄出来了, 哎~俺的50分~~~
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galois_godel 2003-10-17
btw:我记得楼主不是一向做zju的吗?
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galois_godel 2003-10-17
可以到O(n^2)
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galois_godel 2003-10-17
四边形不等式
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LeeMaRS 2003-10-17
楼上的算法复杂度是O(n^3), 呵呵, 再想想有没有什么优化的方法.
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galois_godel 2003-10-17
动态规划

c[i,j]=min(c[i,k]+c[k,j])+len[i,j];
c[1,n]即为所求
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huangry 2003-10-17
我也是两年前提交的 ;)
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