DP：poj1579 Function Run Fun

最新推荐文章于 2018-08-26 19:15:58 发布

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本文介绍了一个三参数递归函数的高效实现方法。通过预先计算并存储递归结果避免重复计算，解决了直接递归导致的大规模计算问题。适用于中等规模的参数值，如a=15, b=15, c=15的情况。

[转]http://blog.youkuaiyun.com/zxy_snow/article/details/6168942

Function Run Fun

Time Limit:

1000ms

Memory limit:

10000kB

题目描述

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.

输入

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.

输出

Print the value for w(a,b,c) for each triple.

样例输入

1 1 1 2 2 2 10 4 6 50 50 50 -1 7 18 -1 -1 -1

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>
using namespace std;
int num[30][30][30];
int w(int a,int b,int c)
{
	if( a <= 0 || b <= 0 || c <= 0 )
		return 1;
	if( a < b && b < c )
		return num[a][b][c-1] + num[a][b-1][c-1] - num[a][b-1][c];
	return num[a-1][b][c] + num[a-1][b-1][c] + num[a-1][b][c-1] - num[a-1][b-1][c-1];
}
int main()
{
	int a,b,c;
	for(a=0; a<=20; a++)
		for(b=0; b<=20; b++)	
			for(c=0; c<=20; c++)
				num[a][b][c] = w(a,b,c) ;
	while( scanf("%d%d%d",&a,&b,&c) != EOF )
	{
		if( a == b && b == c && c == -1 )
			break;
		if( a <= 0 || b <= 0 || c <= 0 )
		{
			printf("w(%d, %d, %d) = 1/n",a,b,c);
			continue;
		}
		if( a > 20 || b > 20 || c > 20 )
		{
			printf("w(%d, %d, %d) = %d/n",a,b,c,num[20][20][20]);
			continue;
		}
		printf("w(%d, %d, %d) = %d/n",a,b,c,num[a][b][c]);
	}
return 0;
}