看来以前题目的数据规模可能都不大，在等报错和超时的时候居然蹦出来100分。
虽然使用的是数组，但是还是觉得为了避免太多的数据移动，用链表实现可能更好一点。

import java.util.*;
public class Main {
public static void main(String[] args) {
	Scanner in = new Scanner(System.in);
	int N = in.nextInt();
	int M = in.nextInt();
	int[][] windows = new int[N][5];
	String[] s = new String[N];
	int num =0;
for(int i =0;i<N;i++) {
	windows[i][0] = in.nextInt();//x1
	windows[i][1] = in.nextInt();//y1
	windows[i][2] = in.nextInt();//x2
	windows[i][3] = in.nextInt();//y2
	windows[i][4] = i+1;//窗口编号
}
for(int i =0;i<M;i++) {
	int x = in.nextInt();
	int y = in.nextInt();
	boolean flag = false;
	for(int j =N-1;j>=0;j--) {//向上遍历
		if(x>=windows[j][0]&& x <=windows[j][2] && y>=windows[j][1]&& y<=windows[j][3] ) {//点中了窗口
					System.out.println(windows[j][4]);
			flag = true;
			if(j == N-1) break;	
			//移动窗口位置
			int x1 = windows[j][0];
			int y1 = windows[j][1];
			int x2 = windows[j][2];
			int y2 = windows[j][3];
			int number = windows[j][4];
			for(int p = j;p<N-1;p++) {
				windows[p][0] = windows[p+1][0];
				windows[p][1] = windows[p+1][1];
				windows[p][2] = windows[p+1][2];
				windows[p][3] = windows[p+1][3];
				windows[p][4] = windows[p+1][4];
			}
			windows[N-1][0] = x1;
			windows[N-1][1] = y1;
			windows[N-1][2] = x2;
			windows[N-1][3] = y2;
			windows[N-1][4] = number;
			break;
		}
	}
	if(!flag) {	
		System.out.println("IGNORED");
	}
	
}//for
	in.close();
}

}