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\n", w(a, b, c));

    }

    return 0;

}