Function Run Fun一个简单的DP问题分析

fuliang

<div class="prob_text">问题:
We all love recursion! Don't we?

Consider a three-parameter recursive function w(a, b, c):

if a <= 0 or b <= 0 or c <= 0, then w(a, b, c) returns:
1

if a > 20 or b > 20 or c > 20, then w(a, b, c) returns:
w(20, 20, 20)

if a < b and b < c, then w(a, b, c) returns:
w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c)

otherwise it returns:
w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1)

This is an easy function to implement. The problem is, if implemented directly, for moderate values of a, b and c (for example, a = 15, b = 15, c = 15), the program takes hours to run because of the massive recursion.


Input

The input for your program will be a series of integer triples, one per line, until the end-of-file flag of -1 -1 -1. Using the above technique, you are to calculate w(a, b, c) efficiently and print the result. For example:
1 1 12 2 210 4 650 50 50-1 7 18-1 -1 -1Output

Print the value for w(a,b,c) for each triple, like this:w(1, 1, 1) = 2w(2, 2, 2) = 4w(10, 4, 6) = 523w(50, 50, 50) = 1048576w(-1, 7, 18) = 1分析:这个问题直接递归写，效率相当低，当 a = 15, b = 15, c = 15，就得运行好几个小时才能出来结果。这个问题其实应该是最简单的一种动态规划题目了，因为递推式已经给出了。而动态规划的难点
在于构造出递推式。当时熟悉用程序求解递推式，也是动态规划的三个步骤之一（最优解结
构分析、构造递推式、自底向上求解递推式[也可以递归的lookup式的]）
此题只需要从底向上求解下面递推式就行：
w(a,b,c) = if a <= 0 or b <= 0 or c <= 0 return 1;
           else if a > 20 or b > 20 or c > 20  return w(20, 20, 20);
           else if a < b and b < c, then w(a, b, c) w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c);
           else return w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1)
写一个方便的函数：<div class="dp-highlighter"><div class="bar" />

• int va(int i,int j,int k){     
  
•    if(i <= 0 || j <= 0 || k <= 0)   return 1;     
  
•   return a[j][k];   
  
• }