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;
}