Mondriaan's Dream

题目链接：http://poj.org/problem?id=2411

题目：

Mondriaan's Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 15360		Accepted: 8861

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

题目大意：给你一个已知长和宽的房间，让你用1*2和2*1的地板将其铺满，问有多少种方法可以实现

解题思路：

当在铺第i行时，需要考虑到第i-1行的情况还有当前行地板之间的距离，因此f[i][j]数组代表的就是当第i-1行已经铺满的情况下，第i行，用的情况为j覆盖的方案数，之后需要考虑的就是铺地板时需要考虑的三种情况，分别是：横着放，那么是因为上一行当前列及下一列没有空位，竖着放，那么上一行当前列就会空出空位，不放，那么就是上一行当前列没有位置。具体看代码

<span style="font-size:18px;">#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int n,m;
int i;
long long f[12][1<<12];///在前i-1行已经覆盖满的情况下，第i行用s情况覆盖的方案数
void dfs(int p,int now,int last)
{
    if(p>m) return;
    if(p==m)
        f[i][now]+=f[i-1][last];

    dfs(p+1,(now<<1)|1,(last<<1));///竖着放，会占据到当前列的上一行，因此通过左移让上一列的那个位置空出来
    dfs(p+1,(now<<1),(last<<1)|1);///不放，上一行当前的一列就为1
    dfs(p+2,(now<<2)|3,(last<<2)|3);///横着放，因为上一行当前列没有位置
}
int main()
{
    while(scanf("%d%d",&n,&m)&&n&&m)
    {
        if(n%2&&m%2)///排除行列都为奇数的情况
        {
            printf("0\n");
            continue;
        }
        if(m>n)///让小的作为列
            swap(n,m);
        memset(f,0,sizeof(f));
        f[0][(1<<m)-1]=1;///第0行全部覆盖的方案数为1
        for( i=1;i<=n;i++)
        {
            dfs(0,0,0);
        }
        cout<<f[n][(1<<m)-1]<<endl;
    }

}
</span>