HDU 2873 Bomb Game

题目链接：HDU 2873【Bomb Game】

思路

       数据范围较小，直接暴力求所有状态的SG值，然后将棋盘上所有炸弹的对应位置的SG值异或起来就可以得到当前局面的结果。对于相同位置的上有两个炸弹会自动爆炸，本来他们的SG值的异或和就为0，所以可以不用管。

代码

int n, m, vis[N * N], sg[N][N], test;
char mp[N][N];
void SG() {
  memset(sg, 0, sizeof sg);
  memset(vis, 0, sizeof vis);
  for (int i = 1; i <= 50; i++) {
    sg[i][1] = sg[1][i] = i - 1;
  }

  for (int i = 2; i <= 50; i++) {
    for (int j = 2; j <= 50; j++) {
      for (int a = 1; a < i; a++) {
        for (int b = 1; b < j; b++) {
          vis[sg[i][b] ^ sg[a][j]] = i * 100 + j;
        }
      }
      int temp = i * 100 + j;
      while (vis[sg[i][j]] == temp) {
        sg[i][j]++;
      }
    }
  }
}

void solve() {
  while (~scanf("%d%d", &n, &m)) {
    if (n == 0 || m == 0) {
      break;
    }
    ll res = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%s", mp[i] + 1);
      for (int j = 1; j <= m; j++) {
        if (mp[i][j] == '#')
          res ^= sg[i][j];
      }
    }
    if (res == 0) {
      cout << "Jack" << endl;
    } else {
      cout << "John" << endl;
    }
  }
}

int main() {
  SG();
  solve();
  return 0;
}