HDU 1331 Function Run Fun（记忆化搜索）

Posted 2020-07-12 Annetree的博客

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

 

Source

Pacific Northwest 1999

 

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以下是我直接递归的错误代码。

 

#include<stdio.h>
int w(int a,int b,int c)
{
    if(a<=0||b<=0||c<=0)return 1;
    else if(a>20||b>20||c>20)return w(20,20,20);
    else if(a<b&&b<c)return w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
    else return 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;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        if(a==-1&&b==-1&&c==-1)break;
        printf("w(%d, %d, %d) = %d\\n",a,b,c,w(a,b,c));
    }
}

 

运行结果是这样的。

然后我发现在递归的过程中有很多重复的部分，导致超时，所以要用记忆化搜索来解决。

也就是把已经计算出来的结果放在一个数组里保存，下次计算到这里的时候直接读取结果就可以了。

下面是正确代码。

#include<stdio.h>
int dp[20+5][20+5][20+5]={0};
int w(int a,int b,int c)
{
    if(a<=0||b<=0||c<=0)return 1;
    if(dp[a][b][c]!=0)return dp[a][b][c];//直接读取
    else if(a<b&&b<c)
    {
        dp[a][b][c]=w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
        return dp[a][b][c];
    }
    else 
    {
        dp[a][b][c]= w(a-1,b,c)+w(a-1,b-1,c)+w(a-1,b,c-1)-w(a-1,b-1,c-1);
        return dp[a][b][c];
    }
}
int main()
{
    int a,b,c,res;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        if(a==-1&&b==-1&&c==-1)break;
        if(a<=0||b<=0||c<=0)res=1;
        else if(a>20||b>20||c>20)res=w(20,20,20);
        else res=w(a,b,c);
        printf("w(%d, %d, %d) = %d\\n",a,b,c,res);
    }
}