POJ2411——状态压缩+DFS—

POJ2411——状态压缩+DFS——Mondriaan's Dream

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

Sample Output

1
0
1
2
3
5
144
51205
大意：给你行数和列数，问你总共有多少种铺地的方法，状态转移一共三种情况
（我们把竖的放的上面一个看做是0下面看做是1）
1，now&(1<<(m-1-step))) == 0    pre<<1|1  不铺
2. now&(1<<(m-2-step))) != 0   pre<<2|3   横着铺
3. 其他情况都竖着铺   pre<<1
dfs遍历所有的情况   
0    1 1   1
1    1 1   0
当step == m 的时候说明全部铺满了 当前dp值只要加上上一行的dp值
被第二种坑了。。如果now&(1<<(m-2-step)) == 1 的话并不是所有的情况，因为二进制会出现11....

#include<cstdio>
#include<cstring>
#include<algorithm>
long long dp[12][2148];
int n,m;
void dfs(int step,int pre,int now,int row)
{
    if(step == m){
        dp[row][now] += dp[row-1][pre];
        return ;
    }
    if((now & (1 << (m - step - 1))) == 0)
        dfs(step+1,pre<<1|1,now,row);
    else {
        if(step + 1 == m) dfs(step+1,pre<<1,now,row);
        else {
            if(now & (1 << (m - step - 2)))
                dfs(step+2,pre<<2|3,now,row);
            dfs(step+1,pre<<1,now,row);
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        if( n == 0 && m == 0)
            break;
        memset(dp,0,sizeof(dp));
        dp[0][(1<<m)-1] = 1;
        for(int i = 1; i <= n ; i++)
            for(int j = 0 ;j < (1 << m); j++)
                dfs(0,0,j,i);
        printf("%lld
",dp[n][(1<<m)-1]);
    }
    return 0;
}