Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 4.

构造一个新的矩阵dp，dp[i][j]表示以点(i, j)为右下角的正方形的边长；状态转移方程：

dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1])) + 1;//注意min

对于题目所给的例子就有：

1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0

转化成：

1 0 1 0 0 1 0 1 1 1 1 1 1 2 1 1 0 0 1 0

class Solution {
public:
  int maximalSquare(vector<vector<char>>& matrix) {
    if (matrix.empty() || matrix[0].empty()) return 0;
    int r = matrix.size(),c = matrix[0].size();
    vector<vector<int>>dp(r, vector<int>(c, 0));
    int res = 0;
    for (int i = 0; i < r; i++){
      if (matrix[i][0]=='1'){
        dp[i][0] = 1;
        res = 1;
      }
    }
    for (int j = 0; j < c; j++){
      if (matrix[0][j] == '1'){
        dp[0][j] = 1;
        res = 1;
      }
    }
    for (int i = 1; i < r; i++){
      for (int j = 1; j < c; j++){
        if (matrix[i][j] == '1'){
          dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
          if (dp[i][j]>res)
            res = dp[i][j];
        }
      }
    }
    return res*res;
  }
};