【题解】poj2411 Mondriaan's Dream 状压DP

最新推荐文章于 2021-08-23 11:19:49 发布

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#状压DP

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这篇博客介绍了如何解决一道关于填充小矩形到大矩形的问题，其中大矩形的尺寸为整数。重点讨论了使用状态压缩动态规划（DP）的方法来计算不同填充方式的数量，特别提到了状态转移方程，并给出了样例输入和输出。文章以实际的代码和思路总结，帮助读者理解状压DP的应用。

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题目链接

Description

Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his ‘toilet series’ (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and rectangles), he dreamt of filling a large rectangle with small rectangles of width 2 and height 1 in varying ways.
在这里插入图片描述

Expert as he was in this material, he saw at a glance that he’ll need a computer to calculate the number of ways to fill the large rectangle whose dimensions were integer values, as well. Help him, so that his dream won’t turn into a nightmare!

Input

The input contains several test cases. Each test case is made up of two integer numbers: the height h and the width w of the large rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.

Output

For each test case, output the number of different ways the given rectangle can be filled with small rectangles of size 2 times 1. Assume the given large rectangle is oriented, i.e. count symmetrical tilings multiple times.

Sample Input

1 2
1 3
1 4
2 2
2 3
2 4
2 11
4 11
0 0

Sample Output

1
0
1
2
3
5
144
51205
在这里插入图片描述

设 $F [i, j]$ 表示第 $i$ 行的形态为 $j$ 时，前 $i$ 行分割方案的总数。 $j$ 是用十进制整数记录的 $M$ 位二进制数， $j$ 的第 $k$ 为为1表示是一个竖着的长方形的上半部分。
$F[i,j]=\sum_{j\& k=0且j\mid k\in s}F[i-1,k]$

#include<cstdio>
#include<cstring>
typedef long long ll;
int n,m;
ll f[12][1<<11];
bool S[1<<11];
int main()
{
	//freopen("in.txt","r",stdin);
	while(~scanf("%d%d",&n,&m)&&n)
	{
	    for(int i=0;i<1<<m;i++)//预处理出连续偶数个零的状态 
	    {
		    bool cnt=0,odd=0;//多少个连续的0，奇偶性
		    for(int j=0;j<m;j++)
			    if(i>>j&1)odd|=cnt,cnt=0;
			    else cnt^=1;
		    S[i]=odd|cnt?0:1; 
	    }
	    f[0][0]=1;
		for(int i=1;i<=n;i++)
		    for(int j=0;j<1<<m;j++)
			{	
				f[i][j]=0;
				for(int k=0;k<1<<m;k++)
		            if((j&k)==0&&S[j|k])
		                f[i][j]+=f[i-1][k];
		    }
		printf("%lld\n",f[n][0]);
	}
	return 0;
}

总结

状压DP好题