DFS深度优先搜索(5)--poj1579(简单记忆化搜索)

•                                                                                                      Function Run Fun

                                                       Time Limit:1000MS    Memory Limit:10000KB    64bit IO Format:%lld & %llu

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

                                 

           这道题状态方程都给出来了，直接敲代码就行了

#include 
#include 
int w[25][25][25];
int vis[25][25][25];               //标记是否算过了  算过了就直接return对应的w值就行了(记忆化过程)  
int Dfs(int x,int y,int z)
{
	if(x>0&&x<=20&&y>0&&y<=20&&z>0&&z<=20&&vis[x][y][z])return w[x][y][z];
	if(x>0&&x<=20&&y>0&&y<=20&&z>0&&z<=20)vis[x][y][z]=1;
	if(x<=0||y<=0||z<=0)return 1;
	else if(x>20||y>20||z>20)return Dfs(20,20,20);
	else if(x