POJ-1579 Function Run Fun_poj 1579 c++代码-优快云博客

本文链接：https://blog.youkuaiyun.com/j_sure/article/details/24189811

本文介绍了一种记忆化搜索技术，用于优化解决复杂递归问题的效率。通过案例分析，展示如何通过记忆化避免重复计算，显著提高程序运行速度。重点在于理解记忆化在递归函数实现中的应用，以及如何在输入参数相同的情况下重用计算结果。

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Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K

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

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

————————————————————心乱的分割线————————————————————

思路：最裸的记忆化搜索，裸就裸在直接跟你说，你在哪需要记忆化一下。

对于DFS，重复的计算是相当浪费时间的。这时候就需要利用参数和数组进行记忆化。对DFS来说，参数意味着状态。某一状态如果已经计算，直接返回值就行了。如果并不能确定0能否代表该状态是否已经计算，就要另开一个vis数组。

代码如下：

/****************************************/
 #include <cstdio>
 #include <cstring>
/****************************************/
bool vis[25][25][25];
int ans[25][25][25];
int w(int a, int b, int c) {
	int ret;//小技巧，精简代码
	if(a <= 0 || b <= 0 || c <= 0)
		return 1;
	else if(a > 20 || b > 20 || c > 20)
		return w(20, 20, 20);
	if(vis[a][b][c])//如果存在记忆，直接返回答案
		return ans[a][b][c];
	vis[a][b][c] = true;//记忆化
	if(a < b && b < c) {
		ret = w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c);
	} else {
		ret = w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1);
	}
	return ans[a][b][c] = ret;//返回答案值的同时储存进数组中
}
int main() {
	int a, b, c;
	while(~scanf("%d%d%d", &a, &b, &c), a!=-1||b!=-1||c!=-1) {
		memset(vis, 0, sizeof(vis));
		printf("w(%d, %d, %d) = %d\n", a, b, c, w(a, b, c));
	}
	return 0;
}