• ACM学习历程——HDU1331 Function Run Fun(锻炼多维dp的打表)


    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
     
     
    递推的式子是直接给出来的。里面最关键的两个式子是:
    f[i][j][k] = f[i][j][k-1] + f[i][j-1][k-1] - f[i][j-1][k];
    以及
    f[i][j][k] = f[i-1][j][k] + f[i-1][j-1][k] + f[i-1][j][k-1] - f[i-1][j-1][k-1];
    可知这个三维dp式子,前一个式子都是在i那一维,通过特征观察可知,是j一层一层递推的(或者k)。
    而第二个式子又可以看出是i一层层的递推。
    故递推的时候只需要三个for循环就搞定。
    但是还需要注意的是
    任意i,j,f[i][j][0] = f[i][0][j] = f[0][i][j] = 1;
    这个条件给每一维的0这个面都赋成了1,有了这个初始化,就可以放心地递推了。
    代码:
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <set>
    #include <map>
    #include <vector>
    #include <queue>
    #include <string>
    #define inf 0x3fffffff
    #define eps 1e-10
    
    using namespace std;
    
    int f[25][25][25];
    
    void Init()
    {
        for (int i = 0; i <= 20; i++)
            for (int j = 0; j <= 20; j++)
                f[i][j][0] = f[i][0][j] = f[0][i][j] = 1;
        for (int i = 1; i <= 20; i++)
            for (int j = 1; j <= 20; j++)
                for (int k = 1; k <= 20; k++)
                {
                    if (i < j && j < k)
                        f[i][j][k] = f[i][j][k-1] + f[i][j-1][k-1] - f[i][j-1][k];
                    else
                        f[i][j][k] = f[i-1][j][k] + f[i-1][j-1][k] + f[i-1][j][k-1] - f[i-1][j-1][k-1];
                }
    }
    
    int w(int a, int b, int c)
    {
        if (a <= 0 || b <= 0 || c <= 0)
            return 1;
        if (a > 20 || b > 20 || c > 20)
            return f[20][20][20];
        return f[a][b][c];
    }
    
    int main()
    {
        //freopen("test.txt", "r", stdin);
        Init();
        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
    ", w(a, b, c));
        }
        return 0;
    }
    
     
     
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  • 原文地址:https://www.cnblogs.com/andyqsmart/p/4111449.html
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