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  1. #include <iostream>
  2. #include <cstring>
  3. #include <algorithm>
  4. using namespace std;
  5.  
  6. const int N = 22;
  7. const int INF = 1e9;
  8.  
  9. int n;
  10. int a[N][N];   // 初始矩阵
  11. int x[N][N];   // 主动翻转标记
  12.  
  13. int main() {
  14.     cin >> n;
  15.     for (int i = 1; i <= n; ++i) {
  16.         string s;
  17.         cin >> s;
  18.         for (int j = 1; j <= n; ++j) {
  19.             a[i][j] = s[j - 1] - '0';
  20.         }
  21.     }
  22.  
  23.     int ans = INF;
  24.     // 枚举第一行的所有状态，共 2^n 种
  25.     for (int mask = 0; mask < (1 << n); ++mask) {
  26.         memset(x, 0, sizeof x);
  27.         // 设置第一行的翻转情况
  28.         for (int j = 1; j <= n; ++j) {
  29.             x[1][j] = (mask >> (j - 1)) & 1;
  30.         }
  31.  
  32.         // 递推计算第 2 ~ n 行的翻转情况
  33.         for (int i = 1; i < n; ++i) {
  34.             for (int j = 1; j <= n; ++j) {
  35.                 // 根据第 i 行的方程解出 x[i+1][j]
  36.                 x[i + 1][j] = a[i][j] ^ x[i][j] ^ x[i - 1][j] ^ x[i][j - 1] ^ x[i][j + 1];
  37.             }
  38.         }
  39.  
  40.         // 检查最后一行是否满足全 0
  41.         bool ok = true;
  42.         for (int j = 1; j <= n; ++j) {
  43.             int val = a[n][j] ^ x[n][j] ^ x[n - 1][j] ^ x[n][j - 1] ^ x[n][j + 1];
  44.             if (val != 0) {
  45.                 ok = false;
  46.                 break;
  47.             }
  48.         }
  49.  
  50.         if (ok) {
  51.             int sum = 0;
  52.             for (int i = 1; i <= n; ++i)
  53.                 for (int j = 1; j <= n; ++j)
  54.                     sum += x[i][j];
  55.             ans = min(ans, sum);
  56.         }
  57.     }
  58.  
  59.     if (ans == INF) cout << -1 << endl;
  60.     else cout << ans << endl;
  61.  
  62.     return 0;
  63. }
Success #stdin #stdout 0s 5320KB
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stdin
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3
1 1 1
1 1 1
1 1 1
stdout
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-1
https://ideone.com/kZVXkQ
language:
C++14 (gcc 8.3)
created:
1 day ago
visibility:
public

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