POJ 1579

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
题意：大家爱递归！。说实话我怎么爱递归。简单来讲。根据题目给的要求对一组数字进行递归。常规的dfs会超时，所以进行记忆化搜索。没啥子好说的。

    
    
    
    

     
     
     
     
 
      
#include<stdio.h>
#include<string.h>
int dp[100][100][100];
int dfs(int x,int y,int z)
{ 
	if(x<=0||y<=0||z<=0)                     //判断条件，根据题意来就行了，
	{
	    return 1;
	}
	if(dp[x][y][z]>0)                    //如果已经找到，。直接return；
		return dp[x][y][z];
   if(x>20||y>20||z>20)
	{
		return dfs(20,20,20);
	}
	else if(x<y&&y<z)
	{
	  dp[x][y][z]=dfs(x,y,z-1)+dfs(x,y-1,z-1)-dfs(x,y-1,z);
	}
	else 
	{

	 dp[x][y][z]=dfs(x-1,y,z)+dfs(x-1,y-1,z)+dfs(x-1,y,z-1)-dfs(x-1,y-1,z-1);
	}
	return dp[x][y][z];            
}
int main()
{
	int a,b,c;
	while(scanf("%d%d%d",&a,&b,&c)!=EOF)
	{
		if(a==-1&&b==-1&&c==-1)
			break;
		int i,j,k;
		for(i=0;i<=21;i++)
			for(j=0;j<=21;j++)
				for(k=0;k<=21;k++)
					dp[i][j][k]=0;
		printf("w(%d, %d, %d) = %d\n",a,b,c,dfs(a,b,c));
	}
	return 0;
}