HDU-1331-Function Run Fun（记忆化搜索，dp）

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3921    Accepted Submission(s): 1917

Problem 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

 

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题解：递归代码虽然看起来很简单，但是用时是非常长的，所以为了避免不必要的重复计算，使用记忆化递归。

如果直接用递归做肯定会超时，因为它的子问题实在是太多了，递归过程有许多重复计算的部分，
解决办法就是：用记忆化搜索解决，自底向上的递推，将已求解出来的结果存到数组里保存下来，下次再递归到此处直接读取答案就行，避免重复计算。复杂度O(n^3)。

#include<cstdio>#include<cstring>int dp[22][22][22];int w(int a,int b,int c){if(a<=0||b<=0||c<=0)return 1;if(a>20||b>20||c>20)return w(20,20,20);if(dp[a][b][c])return dp[a][b][c];else 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);elsereturn 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);return dp[a][b][c];}int main(){int a,b,c;while(~scanf("%d%d%d",&a,&b,&c)){memset(dp,0,sizeof(dp));if(a==-1&&b==-1&&c==-1)break;printf("w(%d, %d, %d) = %d\n",a,b,c,w(a,b,c));}return 0;}