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

2024-08-03 10:48:47   220人阅读

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-10using 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;}

 

 
 

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


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