PKU1579 Function Run Fun （DP水题）

/*
 


Function Run Fun
Time Limit: 1000MS        Memory Limit: 10000K
Total Submissions: 6329        Accepted: 3475

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>
using namespace std;
int num[21][21][21];

void init()
{
    int i,j,k;
    for(i = 0;i <= 20;i++)
    {
    for(j = 0;j <= 20;j++)
    {
        for(k = 0;k <= 20;k++)
        {
        if(i == 0 || j == 0 || k == 0)
        {
            num[i][j][k] = 1;
        }
        else if(i < j && j < k)
        {
            num[i][j][k] = num[i][j][k - 1] + num[i][j - 1][k - 1] - num[i][j - 1][k];
        }
        else
        {
            num[i][j][k] = num[i - 1][j][k] + num[i - 1][j - 1][k] + num[i - 1][j][k - 1] - num[i - 1][j - 1][k - 1];
        }
        }
    }
    }
   
           


    return;
   
}

int main(void)
{
    int a,b,c;
    init();
   
    while(1)
    {
    cin>>a>>b>>c;
    if(a == -1 && b == -1 && c == -1)
    {
        break;
       
    }
    else if(a < 0 || b < 0 || c < 0)
    {
        cout<<"w("<<a<<","<<" "<<b<<","<<" "<<c<<")"<<" "<<"="<<" "<<1<<endl;
       
    }
    else if(a > 20 || b > 20 || c > 20)
    {
        cout<<"w("<<a<<","<<" "<<b<<","<<" "<<c<<")"<<" "<<"="<<" "<<num[20][20][20]<<endl;
    }
    else
    {
        cout<<"w("<<a<<","<<" "<<b<<","<<" "<<c<<")"<<" "<<"="<<" "<<num[a][b][c]<<endl;
       
    }
   
   
    }
   
    return 0;
}

 

DP水题。。。不过做了这题清楚地认识到了递归的坏处。。。。。。