【HDU】1331 - Function Run Fun（记忆化递归）

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Function Run Fun

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3922 Accepted Submission(s): 1918

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

Source

Pacific Northwest 1999

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说白了就是动态规划的核心——记忆化搜索。

代码如下：

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
LL num[22][22][22];
LL solve(int a,int b,int c)
{
	if (a <= 0 || b <= 0 || c <= 0)
		return 1;
	else if (a > 20 || b > 20 || c > 20)
		return solve(20,20,20);
	else if (num[a][b][c] != -1)		//记忆化搜索 
		return num[a][b][c];
	else if (a < b && b < c)
		return (num[a][b][c] = solve(a,b,c-1) + solve(a,b-1,c-1) - solve(a,b-1,c));
	else
		return (num[a][b][c] = solve(a-1,b,c) + solve(a-1,b-1,c) + solve(a-1,b,c-1) - solve(a-1,b-1,c-1));
}
int main()
{
	int a,b,c;
	while (~scanf ("%d %d %d",&a,&b,&c))
	{
		if (a == -1 && b == -1 && c == -1)
			break;
		CLR(num,-1);
		LL ans = solve(a,b,c);
		printf ("w(%d, %d, %d) = ",a,b,c);
		printf ("%lld\n",ans);
	}
	return 0;
}