hdu 1400 Mondriaan's Dream 状态压缩

Problem 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 file 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

将1*2的方块放入n*m的格子，问有几种放法。

状态压缩，将一行压缩到一起，0表示此位置没有方块，1表示此位置已经被占用。

首先预处理一下，求出第一行可能的状态。第x位横放则为11不放则为0。

然后dfs进行搜索，第i-1行的第x位为0，则第i行的x位必为竖放1，若i-1行的x位为1，则第i行的x位有不放0,横放11两种可能。

#include <iostream>
#include <cstring>

using namespace std;

const long long maxn=(1<<11);

long long n,m,i;

long long f[14][maxn];

void init(long long x,long long now)
{
    if (x==m+1) f[1][now]++;
    if (x>m) return;
    init(x+1,now<<1);
    init(x+2,(now<<2)|3);
}

void dfs(long long x,long long lst,long long nxt)
{
    if (x==m+1){
        f[i][nxt]+=f[i-1][lst];
        //cerr<<x<<" "<<lst<<" "<<nxt<<endl;
        //cerr<<f[i][nxt]<<" "<<f[i-1][lst]<<endl;
    }
    //cerr<<i<<" "<<nxt<<" "<<f[i][nxt]<<endl;
    if (x>m) return;
    dfs(x+1,(lst<<1)|1,nxt<<1);
    dfs(x+1,lst<<1,(nxt<<1)|1);
    dfs(x+2,(lst<<2)|3,(nxt<<2)|3);
}

int main()
{
    while (cin>>n>>m)
    {
        if (m==0&&n==0) break;
        if (m>n) swap(m,n);
        memset(f,0,sizeof(f));
        init(1,0);
        //for (int j=0;j<=(1<<m)-1;j++){//if (f[1][j]>0) cerr<<j<<endl;//cerr<<f[1][j]<<endl;}
        for (i=2;i<=n;i++)
        {
            dfs(1,0,0);
            //for (int j=0;j<=(1<<m)-1;j++){cerr<<f[i][j]<<" ";/*if (f[i][j]>0) cerr<<j<<" ";*/}cerr<<endl;
        }
        cout<<f[n][(1<<m)-1]<<endl;
    }
    return 0;
}