poj 1579 Function Run Fun(记忆化搜索)

Function Run Fun

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 18256 Accepted: 9328

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 12 2 210 4 650 50 50-1 7 18-1 -1 -1

Sample Output

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

思路：递归+通过记录消除冗余

代码：
#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int issolved[100][100][100];int solve(int a,int b,int c){if(a<=0||b<=0||c<=0) return 1;if(a>20||b>20||c>20) return solve(20,20,20);if(issolved[a][b][c]) return issolved[a][b][c];if(a<b&&b<c) return issolved[a][b][c]=solve(a,b,c-1)+solve(a,b-1,c-1)-solve(a,b-1,c);else return issolved[a][b][c]=solve(a-1,b,c)+solve(a-1,b-1,c)+solve(a-1,b,c-1)-solve(a-1,b-1,c-1);}int main(){int x,y,z;memset(issolved,0,sizeof(issolved));while(~scanf("%d%d%d",&x,&y,&z)){if(x==-1&&y==-1&&z==-1) break;int ans=solve(x,y,z);printf("w(%d, %d, %d) = %d\n",x,y,z,ans);}return 0;}