POJ 1579-Function Run Fun（内存搜索）-白红宇

POJ 1579-Function Run Fun（内存搜索）

阅读量：5879 次

发布时间：2019-06-19

本文共 2154 字，大约阅读时间需要 7 分钟。

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16503		Accepted: 8514

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:

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

巨水。。

一開始wa了一组数据 （0,21,21）

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                #include 
                 
                  #define ll long long#define maxn 116#define pp pair
                  
                   #define INF 0x3f3f3f3f#define max(x,y) ( ((x) > (y)) ? (x) : (y) )#define min(x,y) ( ((x) > (y)) ?

(y) : (x) ) using namespace std; int n,m,dp[22][22][22]; int dfs(int x,int y,int z) { if(x<=0||y<=0||z<=0) return 1; if(x>20||y>20||z>20) return dfs(20,20,20); if(dp[x][y][z]!=INF) return dp[x][y][z]; dp[x][y][z]=0; if(x<y&&y<z) dp[x][y][z]+=(dfs(x,y,z-1)+dfs(x,y-1,z-1)-dfs(x,y-1,z)); else dp[x][y][z]+=(dfs(x-1,y,z)+dfs(x-1,y-1,z)+dfs(x-1,y,z-1)-dfs(x-1,y-1,z-1)); return dp[x][y][z]; } int main() { int a,b,c; while(scanf("%d%d%d",&a,&b,&c)) { if(a==-1&&b==-1&&c==-1)break; memset(dp,INF,sizeof(dp)); printf("w(%d, %d, %d) = %d\n",a,b,c,dfs(a,b,c)); } return 0; }

你可能感兴趣的文章