Solid Dominoes Tilings (轮廓线dp打表 + 容器)

第一步先打一个表，就是利用轮廓线DP去打一个没有管有没有分界线组合数量的表

#include<bits/stdc++.h>
using namespace std;

const int mod = 1e9 + ;
const int maxn = <<;
int dp[][maxn + ];
int ans[][];

int solve(int n, int m){
    if(n * m %  == ) return ;
    memset(dp, , sizeof(dp));
    dp[][] = ;
    int ing = , ed = ;
    for(int i = ; i < n; i ++){
        for(int j = ; j < m; j ++){
            swap(ing, ed);
            memset(dp[ing], , sizeof(dp[ing]));
            for(int sta = ; sta < (<<m); sta ++){
                if(sta & (<<j)) dp[ing][sta&(~(<<j))] = (dp[ing][sta&(~(<<j))] + dp[ed][sta]) % mod;//无添加
                if((sta & (<<j)) == ) dp[ing][sta | (<<j)] = (dp[ing][sta | (<<j)] + dp[ed][sta]) % mod;//加上2*1（竖放）
                if(((sta & (<<j)) == ) && (j != m - )) dp[ing][sta | (<<(j+))] = (dp[ing][sta | (<<(j+))] + dp[ed][sta]) % mod;//加上1*2（横放）
            }
        }
    }
    return dp[ing][];
}

int main(){
    freopen("data.txt", "w", stdout);
    for(int i = ; i < ; i ++)
        for(int j = ; j < ; j ++)
            ans[i][j] = solve(i, j);
    for(int i = 0; i < ; i ++){
        for(int j = 0; j < ; j ++)printf("%d, ",ans[i][j]);
    }
    return ;
}

然后用容器原理加上枚举列当前分界线情况去递推容斥。具体如代码：

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const LL mod=1e9+;

LL dp[][]={,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,};
LL f[];
int b[],cnt,siz,n,m;
LL solve(int n,int m){
    LL ret = ;
    for(int sta = ; sta < ( << (m - )); sta ++){
        int cnt = , len = ;
        for(int i = ; i < m - ; i ++){
            len ++;
            if(sta >> i & ){
                b[cnt ++] = len;
                len = ;
            }
        }
        b[cnt ++] =  ++len;

        for(int i = ; i <= n; i ++){
            for(int j = ; j < i; j ++){
                LL temp = ;
                for(int k = ; k < cnt; k ++)
                    temp = temp * dp[b[k]][i - j] % mod;
                if(!j) f[i] = temp;
                else f[i] = ( (f[i] - f[j] * temp % mod) % mod + mod ) % mod;
            }
        }
        if(!(cnt&)) ret = ( (ret - f[n]) % mod + mod ) % mod;
        else ret = (ret + f[n]) % mod;
    }
    return ret;
}
int main(){
    while(~scanf("%d%d",&n,&m)) printf("%lld\n",solve(n,m));
    return ;
}