# POJ 1579 Function Run Fun 【记忆化搜索入门】

2022年11月24日
0 收藏 682 点赞 5,183 浏览 1816 个字

## Function Run Fun

 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 20560 Accepted: 10325

### 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 12 2 210 4 650 50 50-1 7 18-1 -1 -1`

Sample Output

`w(1, 1, 1) = 2w(2, 2, 2) = 4w(10, 4, 6) = 523w(50, 50, 50) = 1048576w(-1, 7, 18) = 1`

Source

Pacific Northwest 1999

### 题意概括：

①如果 a < 0 || b < 0 || c < 0；直接返回w( a, b, c )；

②如果 a > 20 || b > 20 || c > 20；返回w( 20, 20, 20 )；

③如果 a < b && b < c ；返回 w(a, b, c-1) + w(a, b-1, c-1) – w(a, b-1, c)；

④ 其他情况返回w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) – w(a-1, b-1, c-1) ；

### 解题思路：

AC code：

` /// POJ 1579 记忆化搜索入门 #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> #define ll long long int #define INF 0x3f3f3f3f using namespace std; const int MAXN = ; int d[MAXN][MAXN][MAXN]; int dfs(int a, int b, int c) {     if(a <=  || b <=  || c <= ) return ;     if(a >  || b >  || c > ) return dfs(, , );     if(d[a][b][c]) return d[a][b][c];     if(a < b && b < c) d[a][b][c] = dfs(a, b, c-)+dfs(a, b-, c-) - dfs(a, b-, c);     else d[a][b][c] = dfs(a-, b, c)+dfs(a-, b-, c)+dfs(a-, b, c-)-dfs(a-,b-,c-);     return d[a][b][c]; } int main() {     int ans, A, B, C;     memset(d, , sizeof(d));     while(~scanf("%d%d%d", &A, &B, &C))     {         if(A == - && B == - && C == -) break;         ans = dfs(A, B, C);         printf("w(%d, %d, %d) = %d\n", A, B, C, ans);     }     return ; }`

python开发_常用的python模块及安装方法

Educational Codeforces Round 11 C. Hard Process 二分
C. Hard Process题目连接：http://www.codeforces.com/contest/660/problem/CDes…

zengkefu@server1:/usr/src\$ uname -aLinux server1 4.10.0-19-generic #21…

Android调用系统相机、自定义相机、处理大图片
Android调用系统相机和自定义相机实例本博文主要是介绍了android上使用相机进行拍照并显示的两种方式，并且由于涉及到要把拍到的照片显…

Struts的使用