//Simplest Task in Windows  http://acm.zju.edu.cn/show_problem.php?pid=2480
//2008.5.15 20:36:21 Accepted 2480 C++ 00:00.00 836K 天将降大任于我
// Please notice that windows may overlap, and the window on top would
//receive the click rather than others 窗口会叠加，并且在顶部的窗口接受点击的几率比其他的还要大
// The windows are given in back-to-front order 窗口是从底到顶地给出的
#include<iostream>
using namespace std;
struct node
{
 int x1,y1,x2,y2;
}a[11];
int main()
{
 int i,n,j,k;int b[51][2];
 while(scanf("%d",&n)&&n)
 {
  for(i=0;i<n;i++)
    scanf("%d%d%d%d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
  scanf("%d",&k);
  for(i=0;i<k;i++)  scanf("%d%d",&b[i][0],&b[i][1]);
         for(i=0;i<k;i++)
   { for(j=n-1;j>=0;j--)
   {
    if(b[i][0]>=a[j].x1&&b[i][0]<=a[j].x2&&b[i][1]>=a[j].y1&&b[i][1]<=a[j].y2)  {printf("%d\n",j);break;}
   }
      if(j<0) printf("-1\n");
         }
 }
 return 0;
}

笨笨来顶贴了