POJ 1579-Function Run Fun（记忆化搜索-递归）

- 日理万妓 2022-06-15 22:22 153阅读 0赞

Function Run Fun

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;18930</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;9612</td> 
  </tr> 
 </tbody> 
</table>

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

Source

[Pacific Northwest 1999][]

## 题目意思： ##

给出a,b,c，按以下递归方式计算w(a, b, c)：

①if a <= 0 or b <= 0 or c <= 0, then w(a, b, c) returns:1；  
②if a > 20 or b > 20 or c > 20, then w(a, b, c) returns:w(20, 20, 20)  
③if 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)  
④otherwise 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)

## 解题思路： ##

直接递归必然超时，所以一定要记忆化搜索。

用dp\[a\]\[b\]\[c\]记录w(a,b,c)的值，每次判断一下如果计算过就不再重复计算，注意防止下标越界的问题，要先处理负数再处理大数。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <bitset>
    #include <map>
    #include <iomanip>
    #include <algorithm>
    #define MAXN 21
    #define INF 0xfffffff
    using namespace std;
    /*
    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)
    */
    int dp[MAXN][MAXN][MAXN];
    int w(int a,int b,int c)
    {
        if(a<=0||b<=0||c<=0) return 1;//为了防止下标越界(为负)所以先处理
        if(a>20||b>20||c>20) return dp[20][20][20]=w(20,20,20);//为了防止下标越界(大于20)所以先处理
        if(dp[a][b][c]!=0) return dp[a][b][c];//已经计算过的不再重复计算
        if(a<b&&b<c)  return dp[a][b][c]=(w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c));
        return dp[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));
    }
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("G:/cbx/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        memset(dp,0,sizeof(dp));
        int a,b,c;
        while(cin>>a>>b>>c)
        {
            if(a==-1&&b==-1&&c==-1) break;
            cout<<"w("<<a<<", "<<b<<", "<<c<<") = "<<w(a,b,c)<<endl;
        }
        return 0;
    }

[Pacific Northwest 1999]: http://poj.org/searchproblem?field=source&key=Pacific+Northwest+1999