真的是个好（毒）题（瘤）。其中枚举的思想尤其值得借鉴。

\(40pts\)：插头\(dp\)，记录插头的同时记录每一列的连接状况，复杂度\(O(N*M*2^{n + m} )\)。

\(100pts\)：容斥\(+\)插头\(/\)轮廓线。目前要维护每两行和每两列的限制，我们把两个限制分开讨论。预处理一下每个子矩阵如果不作限制的可行方案，然后人为地进行限制分割。对于列的限制用容斥解决，对于行的限制套在里面逐行枚举做一次\(dp\)，就可以把复杂度降到可以接受的\(O(m^3*2^n)\)。

代码有借鉴网上。并非完全原创。

\(40pts\)：

#include <bits/stdc++.h>
using namespace std;const int N = 15 + 5;
const int base = 2999830;
const int Mod = 19901013;
const int M = 3000000 + 5;
typedef unsigned int uint;int n, m, cur, las, cnt[2], can[N][N];int nxt[M], head[M], dp[2][M]; uint Hash[2][M];int get_wei (uint zt, int wei) {return (zt >> wei) % 2;
}int alt_wei (uint zt, int wei, int val) {return zt + ((0ll + val - get_wei (zt, wei)) << wei);
}void update (uint zt, int val) {uint _zt = zt % base;for (int i = head[_zt]; i; i = nxt[i]) {if (Hash[cur][i] == zt) {(dp[cur][i] += val) %= Mod; return;}}nxt[++cnt[cur]] = head[_zt];head[_zt] = cnt[cur];Hash[cur][cnt[cur]] = zt;
//  cout << "dp[" << cur << "][" << cnt[cur] << "] = " << val << endl;dp[cur][cnt[cur]] = val;
} int get_val (uint zt) {uint _zt = zt % base;for (int i = head[_zt]; i; i = nxt[i]) {if (Hash[cur][i] == zt) {return dp[cur][i];}}return 0;
}void change_row () {for (int i = 1; i <= cnt[cur]; ++i) {uint &zt = Hash[cur][i];if (get_wei (zt, m + m) == 0) {dp[cur][i] = 0;//到不了下一行 } for (int i = m; i >= 1; --i) {zt = alt_wei (zt, i, get_wei (zt, i - 1));}zt = alt_wei (zt, 0, 0); // 全都向后移一位 zt = alt_wei (zt, m + m, 0); // 到下一行的标记清空 }
}int solve () {update (1 << (m + m), 1);for (int i = 1; i <= n; ++i) {change_row ();for (int j = 1; j <= m; ++j) {
//          cout << "i = " << i << " j = " << j << endl;las = cur, cur ^= 1, cnt[cur] = 0;memset (head, 0, sizeof (head));for (int k = 1; k <= cnt[las]; ++k) {uint zt = Hash[las][k];int b1 = get_wei (zt, j - 1);int b2 = get_wei (zt, j - 0);int val = dp[las][k];if (val == 0) continue; // 不转移的小优化
//              cout << "b1 = " << b1 << " b2 = " << b2 << endl;if (!can[i][j]) {if (!b1 && !b2) {update (zt, val);}} else {if (b1 == 0 && b2 == 0) {if (can[i][j + 1]) {uint _zt = zt;_zt = alt_wei (_zt, j - 0, 1); // 右插头改为 1 _zt = alt_wei (_zt, m + j, 1); // j 列和 j + 1 列相连update (_zt, val); }if (can[i + 1][j]) {uint _zt = zt;_zt = alt_wei (_zt, j - 1, 1); // 下插头改为 1_zt = alt_wei (_zt, m + m, 1); // 和下一行连通 update (_zt, val);}update (zt, val); // 注意你并不需要放满所有没有障碍的格子}if (b1 + b2 == 1) { // 有一个连入的插头 uint _zt = zt;_zt = alt_wei (_zt, j - 0, 0);_zt = alt_wei (_zt, j - 1, 0);update (_zt, val);}}}
//          cout << "las : " << endl;
//          for (int k = 1; k <= cnt[las]; ++k) {
//              cout << "state = " << Hash[las][k] << " val = " << dp[las][k] << endl;
//          }
//          cout << "cur : " << endl;
//          for (int k = 1; k <= cnt[cur]; ++k) {
//              cout << "state = " << Hash[cur][k] << " val = " << dp[cur][k] << endl;
//          }
//          cout << endl;}}uint fin = 0;for (int i = m + 1; i < m + m; ++i) {fin |= (1 << i); } return get_val (fin);
}int main () {cin >> n >> m;for (int i = 1; i <= n; ++i) {for (int j = 1; j <= m; ++j) {int ch = getchar ();while (ch != '.' && ch != 'x') {ch = getchar ();}if (ch == '.') {can[i][j] = true;}}}
//  for (int i = 1; i <= n; ++i) {
//      for (int j = 1; j <= m; ++j) {
//          cout << can[i][j] << " ";
//      }
//      cout << endl;
//  }cout << solve () << endl;
}

\(100pts\)：

#include <bits/stdc++.h>
using namespace std;const int N = 15 + 5;
const int M = 40000 + 5;
const int Mod = 19901013;int n, m, top, ans;
int f[N], num[N], tmp[2][M], dp[N][N][N][N];bool mp[N][N];int ask (int u, int v) {return (u >> v) % 2;
}void work (int w, int u, int v) {int las = 0, tot = (1 << (v - u + 1)) - 1;for (int i = 0; i <= tot; ++i) {tmp[0][i] = 0;}int t2 = 0;for (int i = u; i <= v; ++i) {if (mp[w][i]) {t2 |= (1 << (i - u));} }tmp[0][t2] = 1;for (int i = w + 1; i <= m + 1; ++i) {int zt = 0;for (int j = u, t = 0; j <= v; ++j, ++t) {zt |= (mp[i][j] << t);int cur = las ^ 1;for (int k = 0; k <= tot; ++k) tmp[cur][k] = 0;for (int k = 0; k <= tot; ++k) {if(!tmp[las][k]) continue;int t2 = k;if (ask(t2,t) != mp[i][j]) {t2 ^= (1 << t);}(tmp[cur][t2] += tmp[las][k]) %= Mod;if (ask (k, t)) continue;if (j != v && !ask (k, t + 1)) {t2 = k | (1 << (t + 1));if (ask (t2, t) != mp[i][j]) {t2 ^= (1 << t);}(tmp[cur][t2] += tmp[las][k]) %= Mod;}if(!mp[i][j]) {t2 = k | (1 << t);(tmp[cur][t2] += tmp[las][k]) %= Mod;}}las = cur;}dp[w][u][i - 1][v] = tmp[las][zt];}
}int main () {cin >> m >> n;for(int i = 1; i <= m; ++i) {for (int j = 1; j <= n; ++j) {int ch = getchar ();while (ch != '.' && ch != 'x') {ch = getchar ();}mp[i][j] = (ch == 'x');}}for (int i = 1; i <= n; ++i) {for (int j = i; j <= n; ++j) {for (int k = 1; k <= m; ++k) {work (k, i, j);}}}int tot = (1 << (n - 1)) - 1;for(int i = 0; i <= tot; ++i) {int top = 1;num[top] = 0;for (int j = 0; j < n - 1; ++j) {if (ask (i, j)) num[++top] = j + 1;}num[++top] = n;for (int d = 1; d <= m; ++d) {for (int u = 1; u <= d; ++u) {int t = 1;for(int j = 2; j <= top; ++j) {int p = num[j - 1] + 1, q = num[j];t = (1ll * t * dp[u][p][d][q]) % Mod;}if (u == 1) {f[d] = t;} else {f[d] = (0ll + f[d] - (1ll * t * f[u - 1])) % Mod;}}}(f[m] += Mod) %= Mod; if (top % 2 == 1) {(ans -= f[m]) %= Mod;} else { (ans += f[m]) %= Mod;}}cout << (ans + Mod) % Mod << endl;
}