【原文转载】Dynamic Programming

在网上看到一篇一位外国牛人写的关于动态规划法的文章，现原文转载如下：“

An important part of given problems can be solved with the help of dynamic programming (DP for short). Being able to tackle problems of this type would greatly increase your skill. I will try to help you in understanding how to solve problems using DP. The article is based on examples, because a raw theory is very hard to understand. 

Note: If you're bored reading one section and you already know what's being discussed in it - skip it and go to the next one. 

Introduction (Beginner)

What is a dynamic programming, how can it be described? 

A DP is an algorithmic technique which is usually based on a recurrent formula and one (or some) starting states. A sub-solution of the problem is constructed from previously found ones. DP solutions have a polynomial complexity which assures a much faster running time than other techniques like backtracking, brute-force etc. 

Now let's see the base of DP with the help of an example: 

Given a list of N coins, their values (V₁, V₂, ... , V_N), and the total sum S. Find the minimum number of coins the sum of which is S (we can use as many coins of one type as we want), or report that it's not possible to select coins in such a way that they sum up to S. 

Now let's start constructing a DP solution: 

First of all we need to find a state for which an optimal solution is found and with the help of which we can find the optimal solution for the next state. 

What does a "state" stand for? 

It's a way to describe a situation, a sub-solution for the problem. For example a state would be the solution for sum i, where i≤S. A smaller state than state i would be the solution for any sum j, where j<i. For finding a state i, we need to first find all smaller states j (j<i) . Having found the minimum number of coins which sum up to i, we can easily find the next state - the solution for i+1. 

How can we find it? 

It is simple - for each coin j, V_j≤i, look at the minimum number of coins found for the i-V_jsum (we have already found it previously). Let this number bem. If m+1 is less than the minimum number of coins already found for current sum i, then we write the new result for it. 

For a better understanding let's take this example:
Given coins with values 1, 3, and 5.
And the sum S is set to be 11. 

First of all we mark that for state 0 (sum 0) we have found a solution with a minimum number of 0 coins. We then go to sum 1. First, we mark that we haven't yet found a solution for this one (a value of Infinity would be fine). Then we see that only coin 1 is less than or equal to the current sum. Analyzing it, we see that for sum 1-V₁= 0 we have a solution with 0 coins. Because we add one coin to this solution, we'll have a solution with 1 coin for sum 1. It's the only solution yet found for this sum. We write (save) it. Then we proceed to the next state - sum 2. We again see that the only coin which is less or equal to this sum is the first coin, having a value of 1. The optimal solution found for sum (2-1) = 1 is coin 1. This coin 1 plus the first coin will sum up to 2, and thus make a sum of 2 with the help of only 2 coins. This is the best and only solution for sum 2. Now we proceed to sum 3. We now have 2 coins which are to be analyzed - first and second one, having values of 1 and 3. Let's see the first one. There exists a solution for sum 2 (3 - 1) and therefore we can construct from it a solution for sum 3 by adding the first coin to it. Because the best solution for sum 2 that we found has 2 coins, the new solution for sum 3 will have 3 coins. Now let's take the second coin with value equal to 3. The sum for which this coin needs to be added to make 3 , is 0. We know that sum 0 is made up of 0 coins. Thus we can make a sum of 3 with only one coin - 3. We see that it's better than the previous found solution for sum 3 , which was composed of 3 coins. We update it and mark it as having only 1 coin. The same we do for sum 4, and get a solution of 2 coins - 1+3. And so on. 

Pseudocode: 

Set Min[i] equal to Infinity for all of i
Min[0]=0

For i = 1 to S
For j = 0 to N - 1
   If (Vj<=i AND Min[i-Vj]+1<Min[i])

　 Then Min[i]=Min[i-Vj]+1

Output Min[S]

Here are the solutions found for all sums: 

Sum	Min. nr. of coins	Coin value added to a smaller sum to obtain this sum (it is displayed in brackets)
0	0	-
1	1	1 (0)
2	2	1 (1)
3	1	3 (0)
4	2	1 (3)
5	1	5 (0)
6	2	3 (3)
7	3	1 (6)
8	2	3 (5)
9	3	1 (8)
10	2	5 (5)
11	3	1 (10)

As a result we have found a solution of 3 coins which sum up to 11. 

Additionally, by tracking data about how we got to a certain sum from a previous one, we can find what coins were used in building it. For example: to sum 11 we got by adding the coin with value 1 to a sum of 10. To sum 10 we got from 5. To 5 - from 0. This way we find the coins used: 1, 5 and 5. 

Having understood the basic way a DP is used, we may now see a slightly different approach to it. It involves the change (update) of best solution yet found for a sum i, whenever a better solution for this sum was found. In this case the states aren't calculated consecutively. Let's consider the problem above. Start with having a solution of 0 coins for sum 0. Now let's try to add first coin (with value 1) to all sums already found. If the resulting sum t will be composed of fewer coins than the one previously found - we'll update the solution for it. Then we do the same thing for the second coin, third coin, and so on for the rest of them. For example, we first add coin 1 to sum 0 and get sum 1. Because we haven't yet found a possible way to make a sum of 1 - this is the best solution yet found, and we mark S[1]=1. By adding the same coin to sum 1, we'll get sum 2, thus making S[2]=2. And so on for the first coin. After the first coin is processed, take coin 2 (having a value of 3) and consecutively try to add it to each of the sums already found. Adding it to 0, a sum 3 made up of 1 coin will result. Till now, S[3] has been equal to 3, thus the new solution is better than the previously found one. We update it and mark S[3]=1. After adding the same coin to sum 1, we'll get a sum 4 composed of 2 coins. Previously we found a sum of 4 composed of 4 coins; having now found a better solution we update S[4] to 2. The same thing is done for next sums - each time a better solution is found, the results are updated. 

Elementary

To this point, very simple examples have been discussed. Now let's see how to find a way for passing from one state to another, for harder problems. For that we will introduce a new term called recurrent relation, which makes a connection between a lower and a greater state. 

Let's see how it works: 

Given a sequence of N numbers - A[1] , A[2] , ..., A[N] . Find the length of the longest non-decreasing sequence. 

As described above we must first find how to define a "state" which represents a sub-problem and thus we have to find a solution for it. Note that in most cases the states rely on lower states and are independent from greater states. 

Let's define a state i as being the longest non-decreasing sequence which has its last number A[i] . This state carries only data about the length of this sequence. Note that for i<j the state i is independent from j, i.e. doesn't change when we calculate state j. Let's see now how these states are connected to each other. Having found the solutions for all states lower than i, we may now look for state i. At first we initialize it with a solution of 1, which consists only of the i-th number itself. Now for each j<i let's see if it's possible to pass from it to state i. This is possible only when A[j]≤A[i] , thus keeping (assuring) the sequence non-decreasing. So if S[j] (the solution found for state j) + 1 (number A[i] added to this sequence which ends with number A[j] ) is better than a solution found for i (ie. S[j]+1>S[i] ), we make S[i]=S[j]+1. This way we consecutively find the best solutions for each i, until last state N. 

Let's see what happens for a randomly generated sequence: 5, 3, 4, 8, 6, 7: 

I	The length of the longest non-decreasing sequence of first i numbers	The last sequence i from which we "arrived" to this one
1	1	1 (first number itself)
2	1	2 (second number itself)
3	2	2
4	3	3
5	3	3
6	4	4

”。

原文网址：http://www.topcoder.com/tc?d1=tutorials&d2=dynProg&module=Static