Long Dominoes （状态dp）

Find the number of ways to tile an m*n rectangle with long dominoes -- 3*1 rectangles.

Each domino must be completely within the rectangle, dominoes must not overlap (of course, they may touch each other), each point of the rectangle must be covered.

Input ​

The input contains several cases. Each case stands two integers m and n (1 <= m <= 9, 1 <= n <= 30) in a single line. The input ends up with a case of m = n = 0.

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Output

​

Output the number of ways to tile an m*n rectangle with long dominoes.

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Sample Input

3 3
3 10
0 0

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Sample Output

 
 
2
28

题目大概：

用1*3的矩形把n*m的矩形放满，有多少种方法。

思路：

和第一个状态dp一样的题。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define ll long long
const int le=9;
ll dp[31][1<<le][1<<le];
int w,h;
int pan(int pos,int qian,int qian1)//判断状态是否合法
{
    for(int i=pos;i<pos+3;i++)
    {
        if(!(qian&(1<<i))||!(qian1&(1<<i)))
        return 0;
    }
    return 1;
}
void dfs(int pos,int st,int now,int qian,int qian1,ll sum)
{
    if(pos==w)
    {
        dp[st][now][qian]+=sum;
        return;
    }
    if(pos+3<=w&&pan(pos,qian,qian1))//横放
    {
        dfs(pos+3,st,now|(1<<pos)|(1<<(pos+1))|(1<<(pos+2)),qian,qian1,sum);

    }
    if(!(qian&(1<<pos))&&!(qian1&(1<<pos)))//竖放
    {
        dfs(pos+1,st,now|(1<<pos),qian|(1<<pos),qian1,sum);
    }
    if(qian1&(1<<pos))//不放
    {
        dfs(pos+1,st,now,qian,qian1,sum);
    }
}
int main()
{
    while(~scanf("%d%d",&w,&h)&&(w||h))
    {
        memset(dp,0,sizeof(dp));
         dfs(0,1,0,(1<<w)-1,(1<<w)-1,1);
        for(int i=2;i<=h;i++)//枚举行
        {
            for(int j=0;j<(1<<w);j++)//枚举每一个状态
            {
                for(int k=0;k<(1<<w);k++)//枚举上一行状态
                {
                    if(dp[i-1][j][k])
                    dfs(0,i,0,j,k,dp[i-1][j][k]);
                }
            }
        }
        printf("%lld\n",dp[h][(1<<w)-1][(1<<w)-1]);
    }

    return 0;
}