Description of the topic: Given a matrix of two-dimensional 0,1 strings, find the region of the largest square with only one string.
E.g:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
The answer is 4.
Problem-solving ideas: dynamic planning.
Dp[i][j] is the side length of the largest square with (i, j) as the vertex of the lower right corner.
The matrix is ​​scanned from left to right and top to bottom, when found:
1. (i, j), (i-1, j), (i, j-1) When all three positions are 1, the value of (i, j) can be some size above the left side of this value. The regions form a matrix. E.g:
1 0 1 0 0
1 0 1 1 1
1 1 1 1
Next scan (2, 4) and find that (i, j), (i-1, j), (i, j-1) are all three positions, then the value of (i, j) can be combined with five One 1 constitutes a matrix.
But the problem requires the size of the square, so we must find the minimum value min in dp(i-1, j-1), dp(i-1, j), dp(i, j-1), this minimum The value is the minimum side length of the three squares whose vertices are the lower right corner, and the value of (i, j) is scanned and we find that it can be added to a larger matrix, so the smallest square after this complement The side length must be increased by 1, so let dp(i, j) = min + 1.
The reason for finding the minimum value of dp(i-1, j-1), dp(i-1, j), dp(i, j-1) is because the complement is a matrix, and the width of the matrix Limits the side length of the square.
2. (i, j), (i-1, j), (i, j-1) When one of the three positions is 0, it is definitely impossible to make up a rectangle, so dp(i, j) = 0.
During the scan, the variable res records the length of the side of the largest square found.
The final code is as follows:
Class SoluTIon {
Public:
Int maximalSquare(vector"vector"char" & matrix) {
If(matrix.size() == 0) return 0;
Int row = matrix.size(), col = matrix[0].size();
Int res = 0;
Vector "int" dp(col, 0);
//first row
For(int i = 0;i " col;i++) dp[i] = matrix[0][i] - '0';
Res = max(res, *max_element(dp.begin(), dp.end()));
// dp
For(int i = 1;i "row;i++) {
// first col
Dp[0] = matrix[i][0] - '0';
// other cols
For(int j = 1;j "col;j++) {
If(matrix[i][j] == '1') {
Int temp = min(dp[j], dp[j-1]);
Dp[j] = temp + (matrix[i-temp][j-temp] - '0');
} else {
Dp[j] = 0;
}
}
Res = max(res, *max_element(dp.begin(), dp.end()));
}
Return res * res;
}
};
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