HDU 1331 Function Run Fun(记忆化搜索)

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3358    Accepted Submission(s): 1652

Problem Description

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.

 

Output

Print the value for w(a,b,c) for each triple.

 

Sample Input

   
   
   
   

    
    
    
    1 1 1 2 2 2 10 4 6 50 50 50 -1 7 18 -1 -1 -1 
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    w(1, 1, 1) = 2 w(2, 2, 2) = 4 w(10, 4, 6) = 523 w(50, 50, 50) = 1048576 w(-1, 7, 18) = 1
   
   
   
   

分析：

用一个三维数组来存储一下递归调用的结果，每次用到该数据时直接返回值即可。题目没有给说数据a，b，c的大小，但根据题目中的判断，知道范围是【0,20】；

代码如下：

#include <stdio.h>
#include <string.h>
int dp[25][25][25];

int w(int a,int b,int c)
{
	if(dp[a][b][c])
		return dp[a][b][c];
	if(a<=0 || b<=0 || c<=0)
	{
		dp[a][b][c]=1;
		return 1;
	}
	else if(a>20 || b>20 || c>20)
	{
		dp[a][b][c]=w(20,20,20);
		return dp[a][b][c];
	}
	else if(a<b && b<c)
	{
		dp[a][b][c]=w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c);
		return dp[a][b][c];
	}
	else
	{
		dp[a][b][c]=w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1);
		return dp[a][b][c];
	}
}

int main()
{
	int a,b,c;
	int sum;
	while(scanf("%d %d %d",&a,&b,&c),!(a==-1 && b==-1 && c==-1))
	{
		sum=0;
		memset(dp,0,sizeof(dp));
		if(a<=0 || b<=0 || c<=0)
			sum=1;
		else if(a>20 || b>20 || c>20)
			sum=w(20,20,20);
		else
			sum=w(a,b,c);
		printf("w(%d, %d, %d) = %d\n",a,b,c,sum);		
	}
	
	return 0;	
}