POJ2411Mondriaan's Dream题解动态规划DP

Mondriaan's Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 4978		Accepted: 2873

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.

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

Source

Ulm Local 2000

状态定义等见：

http://blog.youkuaiyun.com/power721/archive/2010/09/16/5889003.aspx

感谢POJ的Sayakiss学习了他的代码，让我对状态压缩DP有了初步的了解。

下面的代码就是学习他的代码的产物，其实基本一样

/* 用2进制的01表示不放还是放 第i行只和i-1行有关 枚举i-1行的每个状态，推出由此状态能达到的i行状态 如果i-1行的出发状态某处未放，必然要在i行放一个竖的方块， 所以我对上一行状态按位取反之后的状态就是放置竖方块的状态。 然后用搜索扫一道在i行放横着的方块的所有可能，并且把这些状态累加上i-1的出发状态的方法数， 如果该方法数为0，直接continue。 举个例子 2 4 1111 1111 状态可以由 1100 0000 0110 0011 1111 0000 0000 0000 0000 0000 i-1行取反 0011 1111 1001 1100 0000 横放都只有一种情况 这五种i-1的状态达到，故2 4 的答案为5 */ #include<cstdio> #include<cstring> long long d[12][1<<12],m,n,M,x; void dfs(int i,int s,int p)//第i行状态为s，位置为p { if(p==m)//i行放满 { d[i][s]+=x; return; } dfs(i,s,p+1);//当前位置不放 if(p+2<=m&&!(s&1<<p)&&!(s&1<<p+1)) dfs(i,s|1<<p|1<<p+1,p+2);//当前位置和右边位置横放一个 } int main() { while(scanf("%lld%lld",&m,&n),m+n) { memset(d,0,sizeof(d)); dfs(x=1,0,0);//第一行 int i,j; for(M=1<<m,i=2;i<=n;i++) for(j=0;j<M;j++) if((x=d[i-1][j]))//上一行状态j能达到 dfs(i,~j&(M-1),0);//32位取反截取m位 printf("%lld/n",d[n][M-1]); } }