HDU1331 Function Run Fun 记忆化搜索

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5557    Accepted Submission(s): 2588

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 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

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll arr[22][22][22];
ll w(int a,int b,int c)
{
    if(a<=0||b<=0||c<=0)
        return 1;
    else if(a>20||b>20||c>20)
        return w(20,20,20);
    else if(arr[a][b][c]!=-1)
        return arr[a][b][c];//已经算过的不再重复计算
//记忆化搜索
    else if(a<b&&b<c)
        return arr[a][b][c]=w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
    else
        return arr[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()
{
    int a,b,c;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF){
        if(a==-1&&b==-1&&c==-1)  break;
        memset(arr,-1,sizeof(arr));//标记没有算过
        printf ("w(%d, %d, %d) = ",a,b,c);
		printf ("%lld\n",w(a,b,c));
    }
    return 0;
}