[ACM_POJ_1579]动态规划入门练习(二)Function Run Fun

蓝飞 蓝飞 | 时间:2012-04-03, Tue | 13,069 views
编程算法 

Function Run Fun

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 12362 Accepted: 6466

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

POJ1579

题目一开始就说了We all love recursion! Don't we?(我们都喜欢递归!不是吗?)但从这道题目的复杂度来看,用递归来解决的话必然TLE,因此还是类似斐波那契数列的思路,用空间换时间。

代码如下:

#include<stdio.h>
int w[21][21][21] = {1};
int getw(int a, int b, int c){
	if(a <= 0 || b <= 0 || c <= 0)
		return 1;
	else if(a > 20 || b > 20 || c > 20)
		return w[20][20][20];
	else
		return w[a][b][c];
	return 1;
}
int main(){
	int a, b, c;
	for(a = 1; a <= 20; ++a){
		for(b = 1; b <= 20; ++b){
			for(c = 1; c <= 20; ++c){
				if(a < b && b < c){
					w[a][b][c] = getw(a, b, c-1) + getw(a, b-1, c-1) - getw(a, b-1, c);
				}else{
					w[a][b][c] = getw(a-1, b, c) + getw(a-1, b-1, c) + getw(a-1, b, c-1) - getw(a-1, b-1, c-1);
				}
			}
		}
	}
	while(scanf("%d%d%d", &a, &b, &c) && (a + 1) || (b + 1) || (c + 1))
		printf("w(%d, %d, %d) = %d\n", a, b, c, getw(a, b, c));
	return 0;
}

话说这是我在北京大学ACM上做的第二道题,第一道题是:动态规划入门(一)The Triangle,两道题都是一次通过,让我对北京大学在线评测系统的印象分顿时提高不少 [偷笑] 。。

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2 条评论 »

  1. Lee Lee

    请问第25部的 (a + 1) || (b + 1) || (c + 1) 有什么用啊?为什么我没有他就不能AC?

    1. 判断是否a, b, c都为-1,如果是则结束程序。