POJ1579.Function Run Fun

最新推荐文章于 2021-02-24 17:38:33 发布

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#POJ #递推 #1579

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本文介绍了一个三参数递归函数的实现与优化方法。通过使用动态规划避免重复计算，显著提高了运行效率。文章提供了完整的源代码及调试过程。

试题请参见: http://poj.org/problem?id=1579

题目概述

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.

解题思路

非常简单的递推, 并且已经给出了状态转移方程.

我们要做的就是Copy it. 但是有一个问题, 我们需要记录每一次计算的结果, 以免重复计算造成TLE.

遇到的问题

我写错了判断条件你敢信!!
(请注意是a <= 0而不是a < 0).

源代码

#include <iostream>

const int MAX_SIZE = 21;
int table[MAX_SIZE][MAX_SIZE][MAX_SIZE] = {0};

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);
    }

    if ( table[a][b][c] == 0 ) {
        if ( a < b && b < c ) {
            table[a][b][c] = w(a, b, c - 1) + w(a, b - 1, c - 1) - w(a, b - 1, c);
        } else {
            table[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 table[a][b][c];
}

int main() {
    int a = 0, b = 0, c = 0;

    while ( std::cin >> a >> b >> c ) {
        if ( a == -1 && b == -1 && c == -1 ) {
            break;
        }
        std::cout << "w(" << a << ", " << b << ", " << c << ")"
                  << " = " << w(a, b, c) << std::endl;
    }
    return 0;
}