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  • HDU 1331 Function Run Fun(记忆化搜索)

    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
     

    Recommend
    Ignatius.L   |   We have carefully selected several similar problems for you:  1074 1078 1160 1208 1355 
    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
    ",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
    ",a,b,c,res);
        }
    }
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  • 原文地址:https://www.cnblogs.com/Annetree/p/5525743.html
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