动态规划专题

leecode 221

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        int n = matrix.size();
        if (n == 0) return 0; // 如果矩阵为空，则直接返回0  
        int m = matrix[0].size();
        vector<vector<int>> ans(n, vector<int>(m, 0)); // 初始化ans为n行m列的二维数组，并全部置为0  
        int maxSideLength = 0;
        // 初始化第一行和第一列  
        for (int i = 0; i < n; i++) {
            ans[i][0] = matrix[i][0] - '0'; // 假设矩阵中的字符是'1'或'0'，直接转换为整数  
            maxSideLength = max(maxSideLength, ans[i][0]);
        }
        for (int j = 0; j < m; j++) {
            ans[0][j] = matrix[0][j] - '0';
            maxSideLength = max(maxSideLength, ans[0][j]);
        }


        for (int i = 1; i < n; i++) {
            for (int j = 1; j < m; j++) {
                if (matrix[i][j] == '1') { // 使用'1'字符进行判断，而不是true  
                    ans[i][j] = 1 + min(min(ans[i - 1][j - 1], ans[i][j - 1]), ans[i - 1][j]);
                    maxSideLength = max(maxSideLength, ans[i][j]);
                }
            }
        }
        return maxSideLength * maxSideLength; // 因为返回的是最大正方形的面积，所以需要乘以边长本身  
    }
};

其实这个题目还可以使用二维前缀和来做

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;

int a[101][101];
int b[101][101];

int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			cin >> a[i][j];
		}
	}
	// 计算前缀和
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			b[i][j] = b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1] + a[i][j];
		}
	}
	int len = 1;
	int ans = 0;
	while (len < min(n, m)) {
		for (int i = len; i <= n; i++) {
			for (int j = len; j <= m; j++) {
				if (b[i][j] - b[i - len][j] - b[i][j - len] + b[i - len][j - len] == len * len) {
					ans = max(ans, len);
				}
			}
		}
		len++;
	}
	cout << ans;
	return 0;
}