题目传送门

sol：状压和动规，把每一行的m个01压缩成一个int

• 状压dp

  #include "bits/stdc++.h"
  using namespace std;
  const int MAXN = 15;
  const int MOD = 1e8;
  int n, m;
  bool mp[MAXN][MAXN];
  vector<int> num[MAXN], dp[MAXN];
  void dfs(int i, int j, int k) {
      if (j > m) {
          num[i].push_back(k);
          dp[i].push_back(0);
          return;
      }
      dfs(i, j + 1, k  << 1);
      if ((k & 1) == 0 && mp[i][j + 1] == 1) dfs(i, j + 1, k << 1 | 1);
  }
  int main() {
      scanf("%d%d", &n, &m);
      for (int i = 1; i <= n; i++) {
          for (int j = 1; j <= m; j++) scanf("%d", &mp[i][j]);
          dfs(i, 0, 0);
      }
      for (int j = 0; j < dp[1].size(); j++) dp[1][j] = 1;
      for (int i = 2; i <= n; i++) {
          for (int j = 0; j < dp[i].size(); j++) {
              for (int k = 0; k < dp[i - 1].size(); k++) {
                  if ((num[i][j] & num[i - 1][k]) == 0)
                  {
                      dp[i][j] = (dp[i][j] + dp[i - 1][k]) % MOD;
                  }
              }
          }
      }
      int ans = 0;
      for (int j = 0; j < dp[n].size(); j++) ans = (ans + dp[n][j]) % MOD;
      printf("%d\n", ans);
      return 0;
  }