1018 Mondriaan's Dream 状压DP-地图型变式

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 链接：https://ac.nowcoder.com/acm/contest/25022/1018
来源：牛客网

题目描述

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!

输入描述:

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≤111\leq h,w\leq 111≤h,w≤11.

输出描述:

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.

示例1

输入

复制 1 2 1 3 1 4 2 2 2 3 2 4 2 11 4 11 0 0

1 2
1 3
1 4
2 2
2 3
2 4
2 11
4 11
0 0

输出

复制 1 0 1 2 3 5 144 51205

1
0
1
2
3
5
144
51205

分析

地图形的状压DP，就得设计01有时候比较简单，但这题就比较复杂

如下表示1*2立柱

0
1

如下表示2*1横杆

11

可以想到，这样表示 第一行的状态 i 和第二行的状态 j 相或 i | j 一定是 (1 <<m) - 1，也就是全都是1

i & j 会消除所有的立柱，怎么让剩下的 1 满足都是横杆的合法情况？

只剩下横杆，但是要考虑到，如果是上上一行的立柱到了上一行，导致上一行某个位置是1，当前这一行这个位置 却是1，而旁边也有两个1，就会有连续三个1 。这种情况就是非法的，

所以要写个函数，看看当前状态是不是都是连续偶数个1 就可以了。

同理初始状态也要把这个考虑进去。第一行的所有满足连续偶数个1的都是合法状态。

设dp[i][j] 表示 i 行 状态 j 的合法方案数

初始状态：if(check(i )  ) dp[1][i] = 1;

状态转移：if(check(j&k) dp[i][j] = dp[i-1][k]

 

//-------------------------代码----------------------------

//#define int ll
const int N = 2e6+10;
ll n,m,dp[14][(1<<11) + 10];

bool check(int x) {
    while(x) {
        int temp = 0;
        while(x & 1) {
            x >>= 1;
            temp ++ ;
        }
        if(temp % 2) return 0;
        x >>= 1;
    }
    return 1;
}

void solve()
{
    ms(dp,0);
    if(n == 1) {
        if( m % 2 ) {
            cout<<0<<endl; 
        } else {
            cout<<1<<endl;
        }rt
    }
    
    fo(i,0,(1<<m)-1) {
        if(check(i))dp[1][i] = 1;
    }
    
    fo(i,2,n+1) {
        fo(j,0,(1<<m) - 1) {
            fo(k,0,(1<<m) - 1) {
                if((j | k) != (1 <<m) - 1) continue;
                if(check(j & k)) dp[i][j] += dp[i-1][k];
            }
        }
    }
    cout<<dp[n][(1<<m) - 1]<<endl;
}

signed main(){
    AC();
    clapping();TLE;

//    int t;cin>>t;while(t -- )
    while(cin>>n>>m,n,m)
    solve();
//    {solve(); }
    return 0;
}

/*样例区


*/

//------------------------------------------------------------

 

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