POJ 1579 Function Run Fun 记忆化递归-优快云博客

本文介绍了一个三参数递归函数w(a,b,c)的高效实现方法。通过使用记忆化递归或者动态规划，解决了直接递归导致的时间效率问题。文章提供了一个具体的C语言实现示例，并展示了如何利用三维数组存储中间结果来避免重复计算。

本题使用记忆话递归即可，也可以用动态规划DP

不过得使用三维数组，空间复杂度比较高

Function Run Fun

Time Limit:1000MS

Memory Limit:10000

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

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

/* Author yan
 *POJ 1579
 *Function Run Fun
*/
#include<stdio.h>
int value[61][61][61];
int w(int a,int b,int c)
{
	if(value[a+10][b+10][c+10]!=-1) return value[a+10][b+10][c+10];
	else
	{
		if(a<=0||b<=0||c<=0) return value[a+10][b+10][c+10]=1;
		else if(a>20||b>20||c>20) return value[a+10][b+10][c+10]=w(20,20,20);
		else if(a<b&&b<c) return value[a+10][b+10][c+10]=w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
		else return value[a+10][b+10][c+10]=w(a-1,b,c)+w(a-1,b-1,c)+w(a-1,b,c-1)-w(a-1,b-1,c-1);
	}
}
int main()
{
	int a,b,c;
	//freopen("input","r",stdin);
	while(scanf("%d %d %d",&a,&b,&c)&&(a!=-1||b!=-1||c!=-1))
	{
		memset(value,-1,sizeof(value));
		//printf("%d %d %d/n",a,b,c);
		printf("w(%d, %d, %d) = ",a,b,c);
		if(a<=0||b<=0||c<=0) printf("%d/n",1);
		else
		{
			w(a,b,c);
			printf("%d/n",value[a+10][b+10][c+10]);
		}
	}
	return 0;
}