问题描述

在某图形操作系统中,有 N 个窗口,每个窗口都是一个两边与坐标轴分别平行的矩形区域。窗口的边界上的点也属于该窗口。窗口之间有层次的区别,在多于一个窗口重叠的区域里,只会显示位于顶层的窗口里的内容。

当你点击屏幕上一个点的时候,你就选择了处于被点击位置的最顶层窗口,并且这个窗口就会被移到所有窗口的最顶层,而剩余的窗口的层次顺序不变。如果你点击的位置不属于任何窗口,则系统会忽略你这次点击。

现在我们希望你写一个程序模拟点击窗口的过程。

输入格式

输入的第一行有两个正整数,即 N 和 M。(1 ≤ N ≤ 10,1 ≤ M ≤ 10)

接下来 N 行按照从最下层到最顶层的顺序给出 N 个窗口的位置。每行包含四个非负整数 x₁, y₁, x₂, y₂,表示该窗口的一对顶点坐标分别为 (x₁, y₁) 和 (x₂, y₂)。保证 x₁ < x₂,y₁ < y₂ 。

接下来 M 行每行包含两个非负整数 x, y,表示一次鼠标点击的坐标。

题目中涉及到的所有点和矩形的顶点的 x, y 坐标分别不超过 2559 和1439。

输出格式

输出包括 M 行,每一行表示一次鼠标点击的结果。如果该次鼠标点击选择了一个窗口,则输出这个窗口的编号(窗口按照输入中的顺序从 1 编号到 N);如果没有,则输出"IGNORED"(不含双引号)。

样例输入

样例输出

2
1
1
IGNORED

样例说明

第一次点击的位置同时属于第 1 和第 2 个窗口,但是由于第 2 个窗口在上面,它被选择并且被置于顶层。

第二次点击的位置只属于第 1 个窗口,因此该次点击选择了此窗口并将其置于顶层。现在的三个窗口的层次关系与初始状态恰好相反了。

第三次点击的位置同时属于三个窗口的范围,但是由于现在第 1 个窗口处于顶层,它被选择。

最后点击的 (0, 5) 不属于任何窗口。

编程语言：Python

解题思想：

1.首先存储各类数组，因为最初的数组已经按照大小层次排序输入，对每个点击坐标分别对数组从高到低进行判断。
2.如果点击成功，则保存该窗口此时的坐标，之后将所有窗口的层次重新交换，最后找出该窗口最初的位置即可

时间复杂度：O(n^2)

IDE代码如下：

N, M = map(int, input().split())  # 输入窗口数和点击的次数
InitialWindow = [[0, 0, 0, 0]] * N  # 最初的窗口坐标列表
Window = [[0, 0, 0, 0]] * N  # 改变过程中的窗口坐标列表
Click = [[0, 0]] * M  # 点击坐标列表
Top = 0  # 定义顶层窗口标号
for i in range(N):
    Window[i] = [int(n) for n in input().split()]  # 将一行的输入赋给窗口坐标列表
    InitialWindow[i] = Window[i]
for i in range(M):
    Click[i] = [int(n) for n in input().split()]  # 将一行的输入赋给点击坐标列表
 
Clicked = [0] * N  # 定义层级关系列表
for i in range(M):
    Judge = False   # 判断是否点击成功
    for j in range(N - 1, -1, -1):
        if Click[i][0] >= Window[j][0] and Click[i][0] <= Window[j][2] and Click[i][1] >= Window[j][1] and Click[i][1] <= Window[j][3]:
            Judge = True
            Top = j
            break
    if Judge == True:  # 存在点击成功的窗口
        Topwindow = Window[Top]  # 暂时存储顶层窗口
        for k in range(Top, N - 1):
            Window[k] = Window[k + 1]  # 依次进行窗口层次替换
        Window[N - 1] = Topwindow  # 覆盖顶层窗口
        for l in range(N):  # 找出该顶层窗口在最初的位置标号并输出
            if Window[N - 1][0] == InitialWindow[l][0] and Window[N - 1][1] == InitialWindow[l][1] and Window[N - 1][
                2] == InitialWindow[l][2] and Window[N - 1][3] == InitialWindow[l][3]:
                print(l + 1)
                break
    else:  # 不存在点击成功的窗口
        print("IGNORED")

< >

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