ACM: 动态规划 poj 1579

最新推荐文章于 2016-05-19 23:28:43 发布

Smile_Benson

最新推荐文章于 2016-05-19 23:28:43 发布

阅读量236

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/smile_benson/article/details/51458076

ACM 专栏收录该内容

295 篇文章

订阅专栏

本文介绍了一个三参数递归函数的问题，并提供了一种通过记忆化搜索优化递归算法的方法，使得对于较大输入值也能快速得到结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Function Run Fun
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

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



题意: 给你一个递归函数, 要求你求出结果. (这题够直接的了)

      要你优化它的函数是结果可以快速得出.



解题思路:

          1.记忆化搜索.



代码:

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 21



int a, b, c;

int dp[MAX][MAX][MAX];



int f(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);

    if(dp[a][b][c] != -1)

        return dp[a][b][c];

    else

    {

        if(a < b && b < c)

        {

            dp[a][b][c] = f(a,b,c-1) + f(a,b-1,c-1) - f(a,b-1,c); 

        }

        else

        {

            dp[a][b][c] = f(a-1,b,c) + f(a-1,b-1,c) + f(a-1,b,c-1) - f(a-1,b-1,c-1);

        }

    }

    return dp[a][b][c];

}



int main()

{

//    freopen("input.txt","r",stdin);

    while(scanf("%d %d %d",&a,&b,&c) != EOF)

    {

        if(a == -1 && b == -1 && c == -1)

            break;



        else

        {

            memset(dp,-1,sizeof(dp));

            printf("w(%d, %d, %d) = %d\n",a,b,c,f(a,b,c));

        }

    }

    return 0;

}