记忆化搜索 hdu 1331

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2586    Accepted Submission(s): 1255

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

 

Source

Pacific Northwest 1999

 

非常明显此题通过模拟来做，递归次数太多一定非常浪费时间。不能过去。

能够通过空间换时间的方法。将计算出的值

存储在数组中。然后将全部可能计算出来的值计算一下子就能够了。直接输出就可以。

//考查知识点：记忆化搜索 就是用数组存储。降低递归函数调用的次数和时间 
//考试当天做出来了。今天又做一遍，居然有点生疏了，(⊙﹏⊙)b 

#include<stdio.h>
int s[22][22][22];
void f()
{
	int i,j,k;
	for(i=1;i<22;++i)
	{
		for(j=1;j<22;++j)
		{
			for(k=1;k<22;++k)
			{
				if(i<j&&j<k)
				{
					if(k-1==0)
					s[i][j][k-1]=s[i][j-1][k-1]=1;
					if(j-1==0)
					s[i][j-1][k-1]=s[i][j-1][k]=1;
					s[i][j][k]=s[i][j][k-1]+s[i][j-1][k-1]-s[i][j-1][k];
					continue;
				}
				if(i==1)
				s[i-1][j][k]=s[i-1][j-1][k]=s[i-1][j][k-1]=s[i-1][j-1][k-1]=1;
				if(j==1)
				s[i-1][j-1][k]=s[i-1][j-1][k-1]=1;
				if(k==1)
				s[i-1][j][k-1]=s[i-1][j-1][k-1]=1;
				s[i][j][k]=s[i-1][j][k]+s[i-1][j-1][k]+s[i-1][j][k-1]-s[i-1][j-1][k-1];
			}
		}
	}
}
int main()
{
	int a,b,c;
	f();
	while(~scanf("%d%d%d",&a,&b,&c),!(a==-1&&b==-1&&c==-1))
	{
		if(a<=0||b<=0||c<=0)
		{
			printf("w(%d, %d, %d) = 1
",a,b,c);
			continue;
		}
		if(a>20||b>20||c>20)
		{
			printf("w(%d, %d, %d) = %d
",a,b,c,s[20][20][20]);
			continue;
		}
		printf("w(%d, %d, %d) = %d
",a,b,c,s[a][b][c]);
	}
	return 0;
}