记忆化搜索：POJ1579-Function Run Fun（最基础的记忆化搜索）

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 14815		Accepted: 7659

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

解题心得：

1、这是一个很经典的记忆化搜索，算是记忆化搜索之中最简单的。其实记忆化搜索就是dp，从小到大以此往上转移。就这个题本身来说是一个很很经典的从下到上的递推。

2、题目描述也很清楚，下一个的结果要从上一个的结果中的出，如果下一个的结果未知，那么就会产生大量的递归然后复杂度爆炸，由前到后依次得到答案，上一个答案从下一个中得出，就是动态规划一个很明显的特征。

#include<cstring>
#include<stdio.h>
const int maxn = 25;
struct num
{
    bool flag[maxn][maxn][maxn];
    int Num[maxn][maxn][maxn];
} N;//记录出现过的结果

int w(int a,int b,int c)
{
    //根据题意就好
    if(a <= 0 || b <= 0 || c <= 0)
        return 1;
    if(a > 20 || b > 20 || c > 20)
        return w(20,20,20);
    if(N.flag[a][b][c])
        return N.Num[a][b][c];
    if(a<b && b<c)
        return w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c);

    return 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;

    //预处理一下
    for(int i=1; i<=20; i++)
        for(int j=1; j<=20; j++)
            for(int k=1; k<=20; k++)
            {
                a = w(i,j,k);
                N.flag[i][j][k] = true;
                N.Num[i][j][k] = a;
            }

    while(scanf("%d%d%d",&a,&b,&c))
    {
        if(a == -1 && b == -1 && c == -1)
            break;
        int ans = w(a,b,c);
        printf("w(%d, %d, %d) = %d\n",a,b,c,ans);
    }