hdu1579 Function Run Fun（深搜+记忆化）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?

pid=1579

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

一个典型的自底向上的记忆化搜索问题。地道的DP问题，虽说非常easy，可是确实是太经典了，
题目所给的本身就是一个递归式，可是假设直接用递归做肯定会超时。

解决的方法就是用记忆化搜索，以空间换时间，将已计算好的结果存到数组里以备后用。复杂度O(n^3).

代码一例如以下：

#include <cstdio>
#include <cstring>
int dp[57][57][57];
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(dp[a][b][c])
		return dp[a][b][c];
	if(a < b && b < c)
		dp[a][b][c]=dfs(a,b,c-1)+dfs(a,b-1,c-1)-dfs(a,b-1,c);
	else
		dp[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 dp[a][b][c];
}
int main()
{
	int a, b, c;
	while(~scanf("%d%d%d",&a,&b,&c))
	{
		if(a == -1 && b == -1 && c == -1)
			break;
		int ans = dfs(a,b,c);
		printf("w(%d, %d, %d) = %d
",a,b,c,ans);
	}
	return 0;
}

代码二例如以下：

#include <cstdio>
#define N 20
int a,b,c,w[N+1][N+1][N+1];
int dp(int a,int b,int c)
{
    int i,j,k;
	
    if (a<=0 || b<=0 || c<=0)
        return 1;
    if (a>20 || b>20 || c>20)
        a=b=c=20;
    for (i=0;i<=N;i++)
        for (j=0;j<=N;j++)
            w[0][i][j]=w[i][0][j]=w[i][j][0]=1;
		for (i=1;i<=a;i++)
			for (j=1;j<=b;j++)
				for (k=1;k<=c;k++)
					if (i<j && j<k)
						w[i][j][k]=w[i][j][k-1]+w[i][j-1][k-1]-w[i][j-1][k];
					else
						w[i][j][k]=w[i-1][j][k]+w[i-1][j-1][k]+w[i-1][j][k-1]-w[i-1][j-1][k-1];
					return w[a][b][c];
}

int main()
{
    while (~scanf("%d %d %d",&a,&b,&c))
	{
		if(a == -1 && b == -1 && c == -1)
			break;
		int ans = dp(a,b,c);
        printf("w(%d, %d, %d) = %d
",a,b,c,ans);
	}
    return 0;
}