题目 3180: 蓝桥杯2023年第十四届省赛真题-棋盘

题目 3180: 蓝桥杯2023年第十四届省赛真题-棋盘

题目描述
小蓝拥有 n × n 大小的棋盘，一开始棋盘上全都是白子。小蓝进行了 m 次操作，
每次操作会将棋盘上某个范围内的所有棋子的颜色取反 (也就是白色棋子变为黑色，黑色棋子变为白色)。
请输出所有操作做完后棋盘上每个棋子的颜色。

解题思路
1、两重循环，时间复杂度是$O(n^2)$，优化一下输入输出，能过全部案例

代码

import java.io.*;
import java.util.*;

public class Main {

    static int N = 2000 + 10;
    static int[][] arr = new int[N][N];
    static StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static int next() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }
    

    public static void main(String[] args) throws IOException {
        int n = next();
        int m = next();
        // 棋子颜色，避免弄混了
        int white = 0;
        int black = 1;
        // 行和列坐标
        int x1, y1, x2, y2;
        int t;
        for (int i = 0; i < m; i++) {
            x1 = next() - 1;
            y1 = next() - 1;
            x2 = next() - 1;
            y2 = next() - 1;

            for (int j = x1; j <= x2; j++) {
                for (int k = y1; k <= y2; k++) {
                    t = arr[j][k];
                    t = (t == white) ? black : white;
                    arr[j][k] = t;
                }
            }
        }

        // 打印二维数组
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                out.print(arr[i][j]);
            }
            out.println();
            out.flush();
        }
    }

}

2、使用二维拆分

import java.io.*;
import java.util.*;

public class Main {

    static int N = 2000 + 10;
    static int[][] a = new int[N][N];
    static int[][] b = new int[N][N];
    static StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static int next() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }


    static void insert(int x1, int y1, int x2, int y2, int c) {
        b[x1][y1] += c;
        b[x1][y2 + 1] -= c;
        b[x2 + 1][y1] -= c;
        b[x2 + 1][y2 + 1] += c;
    }

    public static void main(String[] args) throws IOException {
        int n = next();
        int m = next();
        // 行和列坐标
        int x1, y1, x2, y2;
        int t = 1;
        while (m-- > 0) {
            x1 = next();
            y1 = next();
            x2 = next();
            y2 = next();
            // 每次操作均对某一矩阵里面的数+1
            insert(x1, y1, x2, y2, 1);
        }
        // 打印二维数组a
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                a[i][j] = a[i - 1][j] + a[i][j - 1] + a[i - 1][j - 1] + b[i][j];
                // 如果这个位置的数是奇数表示这个位置和最初的颜色相反
                // 如果这个位置的数是偶数，表示操作了偶数次，棋子颜色和原来没操作一样
                if ((a[i][j] & 1) == 1) {
                    out.print(1);
                } else {
                    out.print(0);
                }

            }
            out.println();
            out.flush();
        }
    }

}