经典记忆化搜索

问题

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

题解

直接递归肯定会超时，我们要结合记忆化特点，才有dp操作，防止重复操作，用空间换取时间。时间复杂度O(n^3)。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
#define ms(a, b) memset(a, b, sizeof(a))

int dp[57][57][57];
int dfs(int a,int b,int c)
{
if(a<=0 ||b<=0 ||c<=0)
return 1;
if(a>20 ||b>20 ||c>20)
return dfs(20,20,20);
if(dp[a][b][c])
return dp[a][b][c];
if(a<b && b<c)
dp[a][b][c]=dfs(a,b,c-1)+dfs(a,b-1,c-1)-dfs(a,b-1,c);
else
dp[a][b][c]=dfs(a-1,b,c)+dfs(a-1,b-1,c)+dfs(a-1,b,c-1)-dfs(a-1,b-1,c-1);
return dp[a][b][c];
}
int main()
{
int a,b,c;
while(~scanf("%d%d%d",&a,&b,&c)){
if(a==-1&&b==-1&&c==-1){break;}
int ans=dfs(a,b,c);
printf("w(%d, %d, %d) = %d\n",a,b,c,ans);
}
return 0;
}