【蓝桥杯 | 备赛秘籍】真题解析之差分数组—一维、二维差分一套拿下，附蓝桥杯真题和代码实战！（下）

#include<iostream>
#include<cstdio>
using namespace std;
const int N = 1e3 + 10;
int a[N][N], b[N][N];
# 注意初始化最小要为n+2
void insert(int x1, int y1, int x2, int y2, int c) {
    b[x1][y1] += c;
    b[x2 + 1][y1] -= c;
    b[x1][y2 + 1] -= c;
    b[x2 + 1][y2 + 1] += c;
}

int main() {
    int n, m, q;
    cin >> n >> m >> q;
    
    // 读取矩阵a
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];
    
    // 初始化差分数组b
    // 通过 insert(i, j, i, j, a[i][j]) 将每个元素的值转化为差分数组 b 的初始值。
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            insert(i, j, i, j, a[i][j]);
        }
    }

    // 处理查询
    while (q--) {
        int x1, y1, x2, y2, c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        insert(x1, y1, x2, y2, c);
    }

    // 计算最终的b矩阵
    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]; // 二维前缀和
        }
    }

    // 输出结果
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            printf("%d ", b[i][j]);
        }
        printf("\n");
    }

    return 0;
}

class Solution:
    def solve():
        # 输入棋盘大小 n 和操作次数 m
        n, m = list(map(int, input().split(" ")))
        
        # 构造差分数组，大小为 (n+2) x (n+2)，避免边界问题
        diff = [[0 for _ in range(n+2)] for _ in range(n+2)]
        
        # 处理每个操作，更新差分数组
        for _ in range(m):
            x1, y1, x2, y2 = list(map(int, input().split(" ")))
            # 更新差分数组的四个角
            diff[x1][y1] += 1  # 左上角 +1
            # if y2 + 1 <= n:  # 避免越界
            diff[x1][y2 + 1] -= 1  # 右上角 -1
            # if x2 + 1 <= n:  # 避免越界
            diff[x2 + 1][y1] -= 1  # 左下角 -1
            # if x2 + 1 <= n and y2 + 1 <= n:  # 避免越界 
            diff[x2 + 1][y2 + 1] += 1  # 右下角 +1
        
        # 构造棋盘，存储最终每个位置的颜色
        chess = [[0 for _ in range(n)] for _ in range(n)]
        
        # 通过差分数组计算每个位置的颜色
        for i in range(1, n+1):  # 遍历行
            for j in range(1, n+1):  # 遍历列
                # 计算当前位置的翻转次数
                diff[i][j] += diff[i-1][j] + diff[i][j-1] - diff[i-1][j-1]
                # 根据翻转次数的奇偶性决定颜色
                chess[i-1][j-1] = 0 if diff[i][j] % 2 == 0 else 1 # 对应chess
                # diff[i][j] = 0 if diff[i][j] % 2 == 0 else 1 这种方式要额外处理数组
        
        # 输出棋盘
        # arr = [0 for _ in range(n)] 
        # for i, li in enumerate(diff[1:n+1]):
        #     arr[i] = li[1:n+1]
        # print(arr)
        # 输出棋盘
        for row in chess:
            # 将每行的列表转换为字符串并输出
            print(''.join(map(str, row)))

# 调用 solve 方法
Solution.solve()

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