hdu1579 Function Run Fun 记忆化搜索启蒙题

Function Run Fun

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2116    Accepted Submission(s): 1123

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
   
   
   
   

 

按照题意做就好.  一共20*20*20 =8000个状态.记忆化搜索不重复计算,稳稳过;

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

int w[23][23][23];

int get(int i,int j,int k)
{
	if(i<=0||j<=0||k<=0)  return 1;
    if(i>20||j>20||k>20) return w[20][20][20]=get(20,20,20);
    if(w[i][j][k]!=-1)
        return w[i][j][k];
    if(i<j&&j<k)
    {
        return w[i][j][k]=get(i,j,k-1)+get(i,j-1,k-1)-get(i,j-1,k);
    }
    else
    {
        return w[i][j][k]=get(i-1,j,k)+get(i-1,j-1,k)+get(i-1,j,k-1)-get(i-1,j-1,k-1);
    }
}

int main()
{
    memset(w,-1,sizeof(w));
    int a,b,c;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        if(a==-1&&b==-1&&c==-1)  break;
        printf("w(%d, %d, %d) = ",a,b,c);
        printf("%d\n",get(a,b,c));
    }

    return 0;
}