Codeforces Round #199 (Div. 2) D. Xenia and Dominoes

把 'O' 看成 'X'，然后枚举它的四个方向看看是否能放，然后枚举 $2^4$ 种可能表示每种方向是否放了，放了的话就标成 'X'，就相当于容斥，对于新的图去dp。

dp就是铺地砖，行用二进制来表示是否放了砖块。

#include <bits/stdc++.h>

const int MOD = 1e9 + ;
const int N = 1e4 + ;
const int dir[][] = {{, -}, {, }, {, }, {-, }};
int dp[N][ << ], n, sx, sy;
char s[][N];
int mp1[][N], mp[][N];

void M(int &a) {
    if (a >= MOD) a -= MOD;
    if (a < ) a += MOD;
}

int get(int x) {
    int cnt = ;
    while (x) {
        cnt++;
        x &= (x - );
    }
    return cnt;
}

int DP() {
    for (int i = ; i < ; i++)
        for (int j = ; j <= n; j++)
            dp[j][i] = ;
    dp[][] = ;
    for (int i = ; i <= n; i++) {
        int no = mp[][i] + mp[][i] *  + mp[][i] * ;
        for (int j = ; j < ; j++) {
            if (j & no) continue;
            dp[i][j | no] = dp[i - ][ - j];
            if (j ==  || j == ) {
                M(dp[i][j | no] += dp[i - ][]);
            }
            if (j == ) {
                M(dp[i][j | no] += dp[i - ][]);
                M(dp[i][j | no] += dp[i - ][]);
            }
        }
    }
    return dp[n][];
}

int main() {
    scanf("%d", &n);
    for (int i = ; i < ; i++)
        scanf("%s", s[i] + );
    for (int i = ; i < ; i++)
        for (int j = ; j <= n; j++)
            if (s[i][j] == 'O')
                sx = i, sy = j, mp1[i][j] = ;
            else if (s[i][j] == 'X')
                mp1[i][j] = ;
    std::vector<int> vec;
    for (int i = ; i < ; i++) {
        bool flag = ;
        int x = sx + dir[i][] * , y = sy + dir[i][] * ;
        if (x >=  && x <  && y >=  && y <= n) {
            for (int j = ; j <= ; j++) {
                if (mp1[sx + dir[i][] * j][sy + dir[i][] * j])
                    flag = ;
            }
        } else {
            flag = ;
        }
        if (flag)
            vec.push_back(i);
     }
     int S =  << vec.size();
     int ans = ;
     for (int s0 = ; s0 < S; s0++) {
         memcpy(mp, mp1, sizeof(mp));
         for (int j = ; j < vec.size(); j++) {
             int i = vec[j];
             if (s0 >> j & ) {
                 for (int k = ; k <= ; k++) {
                     int x = sx + dir[i][] * k, y = sy + dir[i][] * k;
                     mp[x][y] = ;
                 }
             }
         }
         int f = get(s0) &  ?  : -;
         M(ans += f * DP());
     }
     printf("%d\n", ans);
     return ;
}