hdu 1331(dp)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1331

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2219    Accepted Submission(s): 1123

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

思路：直接调用递归函数指定超时；所以用dp的思想，这样避免重复的递归调用，水过~~~~~

#include <iostream>#include <stdio.h>#include <string>#include <string.h>#include <cstdio>#include <cmath>using namespace std;int dp[30][30][30];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]>0)return dp[a][b][c];  if(a<b&&b<c)  {   dp[a][b][c]=w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);  }  else  {   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)!=EOF)    {      if(a==-1&&b==-1&&c==-1)break;      memset(dp,0,sizeof(dp));      printf("w(%d, %d, %d) = %d\n",a,b,c,w(a,b,c));    }    return 0;}