HDU-6514 Monitor (+ prefix and a two-dimensional differential)

http://acm.hdu.edu.cn/showproblem.php?pid=6514

 

Problem Description

Xiaoteng has a large area of land for growing crops, and the land can be seen as a rectangle of  n×m. 
But recently Xiaoteng found that his crops were often stolen by a group of people, so he decided to install some monitors to find all the people and then negotiate with them.
However, Xiao Teng bought bad monitors, each monitor can only monitor the crops inside a rectangle. There are p monitors installed by Xiaoteng, and the rectangle monitored by each monitor is known. 
Xiao Teng guess that the thieves would also steal q times of crops. he also guessed the range they were going to steal, which was also a rectangle. Xiao Teng wants to know if his monitors can see all the thieves at a time.

Input

There are mutiple test cases.
Each case starts with a line containing two integers  n,m(1≤n,1≤m,n×m≤107) which represent the area of the land.
And the secend line contain a integer p(1≤p≤106) which represent the number of the monitor Xiaoteng has installed. This is followed by p lines each describing a rectangle. Each of these lines contains four intergers x1,y1,x2 and y2(1≤x1≤x2≤n,1≤y1≤y2≤m) ,meaning the lower left corner and upper right corner of the rectangle.
Next line contain a integer q(1≤q≤106) which represent the number of times that thieves will steal the crops.This is followed by q lines each describing a rectangle. Each of these lines contains four intergers x1,y1,x2 and y2(1≤x1≤x2≤n,1≤y1≤y2≤m),meaning the lower left corner and upper right corner of the rectangle.

Output

For each case you should print  q lines.
Each line containing YES or NO mean the all thieves whether can be seen.

Sample Input

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

Sample Output

YES
NO

Hint

In the picture,the red solid rectangles mean the monitor Xiaoteng installed, and the blue dotted rectangles mean the area will be stolen.

 (X1, y1) is the lower left corner of the rectangle, (x2, y2) for the lower right corner of the rectangle, the positive x direction is vertically upward, the horizontal rightward direction is the positive y

 

 

 

Meaning of the questions:

In a rectangular area of ​​no more than n * m, p-A rectangular, rectangular before q B after A is asked whether all covered.

Ideas:

Since n * m, p, q range is too large, then consider O (n * m + p + q) approach, i.e., the prefix and a two-dimensional differential +.

For Class A rectangle (x1, y1, x2, y2), we need, (x2 + 1, y2 + 1) only at (x1, y1) +1, the (x1, y2 + 1), (x2 + 1 , y1) at -1

After seeking a prefix and the entire area is greater than 0 is the point where the covered class A rectangle.

Where the value is greater than 0 becomes 1, and once again seeking a prefix, the number of points to be covered in a rectangle can be calculated in O (1) time after the handle.
(For details see Notes)

 

Highlights:

Two-dimensional array into a one-dimensional

Two prefix and seek re-assignment

 

code show as below:

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <string>
 5 #include <math.h>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <stack>
 9 #include <queue>
10 #include <set>
11 #include <map>
12 #include <math.h>
13 const int INF=0x3f3f3f3f;
14 typedef long long LL;
15 const  int MOD = + 1E9 . 7 ;
 16  // const Double the PI = ACOS (-1); 
. 17  const  int MAXN 1E7 + = 10 ;
 18 is  the using  namespace STD;
 . 19  
20 is  int A [MAXN]; // two-dimensionally into one-dimensional 
21  int n-, m;
 22 is  
23 is  void the Add ( int X, int Y, int Val) // add tags 
24  {
 25      IF (X> n-|| Y> m)
 26 is          return ;
 27     A [(X- . 1 ) * m + Y] + = Val;
 28  }
 29  
30  int Query ( int X, int Y) // get the value there is mainly used a processing margin 0 
31 is  {
 32      IF (X = = 0 || Y == 0 )
 33 is          return  0 ;
 34 is      return A [(X- . 1 ) * m + Y];
 35  }
 36  
37 [  void the Sum () // find a prefix, and 
38 is  {
 39      for ( int I = . 1 ; i <= n; i ++)
40     {
41         for(int j=1;j<=m;j++)
42         {
43             a[(i-1)*m+j]+=Query(i,j-1)+Query(i-1,j)-Query(i-1,j-1);
44         }
45     }
46 }
47 
48 int main()
49 {
50     while(~scanf("%d %d",&n,&m))
51     {
52         memset(a,0,sizeof(a));
53         int x1,x2,y1,y2;
54         int p,q;
55         scanf("%d",&p);
56         while(p--)
57         {
58             scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
59             Add(x1,y1,1);
60             Add(x2+1,y2+1,1);
61             Add(x1,y2+1,-1);
 62 is              the Add (+ X2 . 1 , Y1, - . 1 );
 63 is          }
 64          the Sum (); // first prefix and dimensional requirements, a [i]> 0 where it is described to cover over 
65          for ( int I = 1 ; I <= n-; I ++) // to point covered reassign 1, facilitating determination of 
66          {
 67              for ( int J = 1 ; J <= m; J ++ )
 68              {
 69                  IF (a [(I- . 1 ) * m + J])
 70                      A [(I- . 1 ) * m + J] = . 1 ;
 71 is             }
 72          }
 73 is          the Sum (); // results and the second prefix request, shall be obtained within the rectangular area stained 
74          Scanf ( " % D " , & Q);
 75          the while (q - )
 76          {
 77              Scanf ( " % D%% D D D% " , & X1, Y1 &, & X2, & Y2);
 78              int ANS = Query (X2, Y2) -Query (X1- . 1 , Y2) -Query (X2, Y1- . 1 ) + Query (X1- . 1 , Y1- . 1 ); // use the prefix and the dyed derived rectangular area 
79              IF (ANS == (X2-X1 + . 1) * (Y2-Y1 + . 1 )) // see stained area is equal to the rectangle total area 
80                  the printf ( " YES \ n- " );
 81              the else 
82                  the printf ( " NO \ n- " );
 83          }
 84      }
 85      return  0 ;
 86 }