POJ 1579 Function Run Fun 【记忆化搜索入门】

题目传送门：http://poj.org/problem?id=1579

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20560		Accepted: 10325

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

题意概括：

要求写一个函数 w( a, b, c) 处理输入数据（多测试）；

①如果 a < 0 || b < 0 || c < 0；直接返回w( a, b, c )；

②如果 a > 20 || b > 20 || c > 20；返回w( 20, 20, 20 )；

③如果 a < b && b < c ；返回 w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c)；

④ 其他情况返回w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1) ；

解题思路：

记忆化搜索裸题。

AC code：

/// POJ 1579 记忆化搜索入门
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#define ll long long int
#define INF 0x3f3f3f3f
using namespace std;

const int MAXN = 23;
int d[MAXN][MAXN][MAXN];

int dfs(int a, int b, int c)
{
    if(a <= 0 || b <= 0 || c <= 0) return 1;
    if(a > 20 || b > 20 || c > 20) return dfs(20, 20, 20);
    if(d[a][b][c]) return d[a][b][c];
    if(a < b && b < c) d[a][b][c] = dfs(a, b, c-1)+dfs(a, b-1, c-1) - dfs(a, b-1, c);
    else d[a][b][c] = dfs(a-1, b, c)+dfs(a-1, b-1, c)+dfs(a-1, b, c-1)-dfs(a-1,b-1,c-1);
    return d[a][b][c];
}
int main()
{
    int ans, A, B, C;
    memset(d, 0, sizeof(d));
    while(~scanf("%d%d%d", &A, &B, &C))
    {
        if(A == -1 && B == -1 && C == -1) break;
        ans = dfs(A, B, C);
        printf("w(%d, %d, %d) = %d
", A, B, C, ans);
    }
    return 0;
}