[Poj2411]Mondriaan's Dream（状压dp）（插头dp）

Mondriaan's Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 18096		Accepted: 10357

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

题意：

给出n * m的棋盘，问用1 * 2的骨牌铺满棋盘的方案数。

分析：

棋盘n，m很小，可以想到状压dp。一般的状压dp是枚举上一维的状态和当前这维状态然后转移。

在蓝书上P384页，也有一种解法。但是网上有另一种做法：http://blog.csdn.net/sf____/article/details/15026397

十分感谢博主的思路。

思路是这样的：

依然定义f[i][j][k]，i为第i行，j为第第j列。k为二进制数，1 - k - 1位为当前行状态，k - m 为上一行状态，当前更新把第k位从上一行更新成当前行状态。

二进制中0表示下一行这个位置可以放数（即当前位置不放或者横着放），1表示下一行这个位置不可以放数（即当前位置竖着放）

可以得到dp状态：

dp[i][j][k ^ (1 << j)] += dp[i][j - 1][k]; -- 1 //竖着放或者不放，因为不可能连续两行不放，所以k ^ （1 << j）和k相同位置必须有一位为1

dp[i][j][k ^ (1 << (j - 1))] += dp[i][j - 1][k]; --2 //从前一格竖着放的转移到当前位置横着放的条件：当前这位上一格必须放了

因为i 和 j其实是刷表的，可以转移成dp[2][k];就可以了

AC代码：

# include <iostream>

# include <cstdio>

# include <cstring>

using namespace std;

const int N =  << ;

long long dp[][N];

int n,m,data;

int main(){

    while(~scanf("%d %d",&m,&n) && (n + m)){

    data = ( << m);

    if(m > n)swap(n,m);

    int now = ;

    memset(dp[now],,sizeof dp[now]);

    dp[now][] = ;

    for(int i = ;i < n;i++){

        for(int j = ;j < m;j++){

            now ^= ;

            memset(dp[now],,sizeof dp[now]);

            for(int k = ;k < data;k++)if(dp[now ^ ][k]){

                dp[now][k ^ ( << j)] += dp[now ^ ][k];

                if(j && (k & ( << (j - ))) && !(k & ( << j)))

                dp[now][k ^ ( << (j - ))] += dp[now ^ ][k];

            }

        }

    }

     printf("%lld\n",dp[now][]);

  }

}

巴特西

[Poj2411]Mondriaan's Dream（状压dp）（插头dp）

题意：

分析：

AC代码：

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