DP(记忆化搜索) -- poj Function Run Fun

Function Run Fun
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 19948 Accepted: 10034
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

#include <iostream>#include <string.h>using namespace std;int dp[21][21][21]; long long Cal(int a,int b,int c){    //如果dp[a][b][c]已经知晓，就不需要再搜索结果     if(dp[a][b][c] != 0)        return dp[a][b][c];    if(a<b && b<c)    {        //保存结果         dp[a][b][c] = Cal(a, b, c-1) + Cal(a, b-1,c-1) - Cal(a, b-1, c);        return dp[a][b][c];    }       else    {        //保存结果         dp[a][b][c] = Cal(a-1, b, c) + Cal(a-1, b-1, c) + Cal(a-1, b, c-1) - Cal(a-1, b-1, c-1);        return dp[a][b][c];    }}int main(){   ios::sync_with_stdio(false);    int a,b,c,aa,bb,cc;    memset(dp,0,sizeof(dp));    for(int i=0; i<=20; ++i)    {        for(int j=0; j<=20; ++j)        {            dp[i][j][0] = 1;            dp[i][0][j] = 1;            dp[0][i][j] = 1;        }    }    while(cin>>a>>b>>c)    {        if(a==-1&&b==-1&&c==-1)            break;        //保留原始数据         aa = a;bb = b; cc = c;        //出现负的，直接打印         if(a<=0 || b<=0 || c<= 0)        {            printf("w(%d, %d, %d) = 1\n",aa,bb,cc);            continue;        }        //存在大于20的，转换一个         if(a> 20 || b >20 || c>20)        {            a=20;b=20;c=20;                 }        Cal(a,b,c);        printf("w(%d, %d, %d) = ",aa,bb,cc);        cout << dp[a][b][c] <<endl;    }   return 0;}