POJ 1579(备忘录算法)

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13491		Accepted: 7023

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

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 100;
int d[N][N][N];
bool vis[N][N][N];

//a,b,c可能是负值，而数组下标无负值，需特殊处理 
int fun(int a,int b,int c)//必须带参数，因为中间需要转存且参数变了 
{
   // if(((a<=0||b<=0||c<=0)||(a>20||b>20||c>20)||(a<b&&b<c))&&!vis[a][b][c])//这个条件不对因为若abc其一为负数便执行了else if，需要把第一个条件单独搞出来 
    if((a<=0||b<=0||c<=0)||(((a>20||b>20||c>20)||(a<b&&b<c))&&!vis[a][b][c]))
    {
        if(a<=0||b<=0||c<=0)
            return 1;
        if(a>20||b>20||c>20&&!vis[a][b][c])
        {
            d[a][b][c] = fun(20,20,20);
            //d[20][20][20] = fun(20,20,20);
            d[20][20][20] = d[a][b][c];
            vis[a][b][c] = 1;
            vis[20][20][20] = 1;
        }
        if(a<b&&b<c&&!vis[a][b][c])
        {
            d[a][b][c-1] = fun(a,b,c-1);
            vis[a][b][c-1] = 1;
            d[a][b-1][c-1] = fun(a,b-1,c-1);
            vis[a][b-1][c-1] = 1;
            d[a][b-1][c] = fun(a,b-1,c);
            vis[a][b-1][c] = 1;
            d[a][b][c] =d[a][b][c-1] + d[a][b-1][c-1] - d[a][b-1][c]; 
            vis[a][b][c] = 1;
        }
    }
    else if(!vis[a][b][c])//加上了if后效率大增 
    {
        d[a-1][b][c] = fun(a-1, b, c);
        vis[a-1][b][c] = 1;
        d[a-1][b-1][c] = fun(a-1, b-1, c);
        vis[a-1][b-1][c] = 1;
        d[a-1][b][c-1] = fun(a-1, b, c-1);
        vis[a-1][b][c-1] = 1;
        d[a-1][b-1][c-1] = fun(a-1, b-1, c-1);
        vis[a-1][b-1][c-1] = 1;
        d[a][b][c] = d[a-1][b][c] + d[a-1][b-1][c] + d[a-1][b][c-1] - d[a-1][b-1][c-1];
        vis[a][b][c] = 1;
    }
    return d[a][b][c]; 
} 

int main()
{
    int i,j,k;
    int a, b, c;
    while(cin>>a>>b>>c&&!(a==-1&&b==-1&&c==-1))
    {
        memset(d,-1,sizeof(d));
        memset(vis,0,sizeof(vis));
        vis[0][0][0] = 1;
        d[0][0][0] = 1;
        int ans = fun(a,b,c);
        //cout<<"w("<<a<<","<<b","<<c<<") = "<<ans<<endl;
        printf("w(%d, %d, %d) = %d\n",a,b,c,ans);//注意有4个空格，原来只注意到了等号两侧的空格吃了一个PE 
    }
    return 0;
}