DFS深度优先搜索(5)--poj1579(简单记忆化搜索)

最新推荐文章于 2021-02-21 22:20:50 发布

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最新推荐文章于 2021-02-21 22:20:50 发布

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分类专栏：【DFS(深度优先搜索)】文章标签： poj1579 记忆化搜索

本文链接：https://blog.youkuaiyun.com/Acmer_Sly/article/details/53440859

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博客介绍了如何运用DFS深度优先搜索解决poj1579问题，通过记忆化搜索优化算法，提高解题效率。

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Function Run Fun

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

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

这道题状态方程都给出来了，直接敲代码就行了

#include <stdio.h>
#include <string.h>
int w[25][25][25];
int vis[25][25][25];               //标记是否算过了  算过了就直接return对应的w值就行了(记忆化过程)  
int Dfs(int x,int y,int z)
{
	if(x>0&&x<=20&&y>0&&y<=20&&z>0&&z<=20&&vis[x][y][z])return w[x][y][z];
	if(x>0&&x<=20&&y>0&&y<=20&&z>0&&z<=20)vis[x][y][z]=1;
	if(x<=0||y<=0||z<=0)return 1;
	else if(x>20||y>20||z>20)return Dfs(20,20,20);
	else if(x<y&&y<z)return Dfs(x,y,z-1)+Dfs(x,y-1,z-1)-Dfs(x,y-1,z);
	else return Dfs(x-1,y,z)+Dfs(x-1,y-1,z)+Dfs(x-1,y,z-1)-Dfs(x-1,y-1,z-1);
}
int main()
{
	int a,c,b,i,j,k;
	memset(vis,0,sizeof(vis));
	for(i=1;i<=20;i++)
	{
		for(j=1;j<=20;j++)
		{
			for(k=1;k<=20;k++)
			{
				w[i][j][k]=Dfs(i,j,k);
			}
		}
	}
	while(scanf("%d%d%d",&a,&b,&c)!=EOF)
	{
		if(a==-1&&b==-1&&c==-1)break;
		printf("w(%d, %d, %d) = %d\n",a,b,c,Dfs(a,b,c));
	}
	return 0;
}