2014-03-2 CCF计算机软件能力认证真题练习：窗口——java实现

题目描述

输入样例

3 4
0 0 4 4
1 1 5 5
2 2 6 6
1 1
0 0
4 4
0 5

输出样例

2
1
1
IGNORED

代码

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        // x[i][0]和x[i][1]分别放x,y，x[i][3]表示这是第几个窗口，便于输出
        int[][] x = new int[n][3];
        int[][] y = new int[n][2];
        for (int i = 0; i < n; i++) {
            x[i][0] = scanner.nextInt();
            y[i][0] = scanner.nextInt();
            x[i][1] = scanner.nextInt();
            y[i][1] = scanner.nextInt();
            x[i][2] = i+1;
        }

        for (int i = 0; i < m; i++) {
            int xi = scanner.nextInt();
            int yi = scanner.nextInt();
            // flag的作用是，如果鼠标点击的位置没有窗口，输出ignored
            boolean flag = true;
            // 由于输入的窗口是从最底层到最顶层的，那么x[n-1]表示最顶层，鼠标点击窗口应该从最顶层往最底层遍历
            for (int j = n-1; j >= 0; j--) {
                if (xi >= x[j][0] && xi <= x[j][1] && yi >= y[j][0] && yi <= y[j][1]){
                    flag = false;
                    System.out.println(x[j][2]);
                    // 为了保持数组中窗口的优先级，需要把当前显示在屏幕最顶层的放到x[n-i]中
                    int[] tempt_x = x[j];
                    int[] tempt_y = y[j];
                    for (int k = j+1; k < n; k++){
                        x[k-1] = x[k];
                        y[k-1] = y[k];
                    }
                    x[n-1] = tempt_x;
                    y[n-1] = tempt_y;
                    break;
                }
            }
            if (flag)
                System.out.println("IGNORED");
        }
    }
}

标签：03,java,scanner,真题,int,nextInt,flag,tempt,窗口
来源： https://blog.csdn.net/a2352159950/article/details/114676890