L. Leverage MDT

题意

给一个 $01$ 矩阵，可以选择一些行翻转。最大化全为 $1$

• 按列枚举
• 对于每一列正反跑两遍单调栈即可

AC代码：

const int N = 1010;
char s[N][N];
int len[N][N];
int up[N], down[N];
int st[N], pos;
int n, m;
int ans;
int main()
{
  sdd(n, m);
  rep(i, 1, n)
    ss(s[i] + 1);
  rep(i, 1, n)
  {
    len[i][m] = 1;
    per(j, m - 1, 1)
    {
      if (s[i][j] != s[i][j + 1])
        len[i][j] = 1;
      else
        len[i][j] = len[i][j + 1] + 1;
    }
  }
  rep(j, 1, m)
  {
    pos = 0;
    st[0] = 0;
    rep(i, 1, n)
    {
      while (len[st[pos]][j] >= len[i][j])
        pos--;
      up[i] = st[pos] + 1;
      st[++pos] = i;
    }
    pos = 0;
    st[0] = n + 1;
    per(i, n, 1)
    {
      while (len[st[pos]][j] >= len[i][j])
        pos--;
      down[i] = st[pos] - 1;
      st[++pos] = i;
    }
    rep(i, 1, n)
    {
      int l = len[i][j], r = down[i] - up[i] + 1;
      ans = max(min(l, r) * min(l, r), ans);
    }
  }
  pd(ans);
  return 0;
}