Monitor（二维数组前缀和）

Monitor

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 163840/163840 K (Java/Others)
Total Submission(s): 872    Accepted Submission(s): 145
 

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.

题意：

n*m的稻田，小腾装了p个监视范围不同区域为矩形的监控，而有q次小偷偷窃，且小偷偷窃的范围给你且也是矩形，问能不能抓到 所有的小偷

#include <bits/stdc++.h>
#define rep1(i,s,e,c) for(int i=s;i<e;i=i+c)
#define rep2(i,s,e,c) for(int i=s;i<=e;i=i+c)
#define rep3(i,e,s,c) for(int i=e;i>=s;i=i-c)
using namespace std;
typedef long long ll;
const int MAX=1e7+1;
const int MOD=1e9+7;
int n,m;
int arr[MAX];//从arr[1]开始
void add(int i,int j,int v){
	if(i>n||j>m) return;//必要,否则会出现不同i,j的值算出的(i-1)*m+j相同的情况 
	arr[(i-1)*m+j]+=v;
}
int query(int i,int j){
	if(i==0||j==0) return 0;
	return arr[(i-1)*m+j];
}
int main()
{
	while(~scanf("%d%d",&n,&m)){
		memset(arr,0,sizeof(arr));
		int p;
		scanf("%d",&p);
		rep1(i,0,p,1){
			int x1,y1,x2,y2;
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			add(x1,y1,1);
			//y2+1列和x2+1行已经不在范围内,所以减1，又因为两次减去，多减去依次(x2+1,y2+1)区域，此处+1 
			add(x2+1,y2+1,1);
			add(x1,y2+1,-1);
			add(x2+1,y2,-1);
		}
		rep2(i,1,n,1){
			rep2(j,1,m,1){
				arr[(i-1)*m+j]=query(i,j-1)+query(i-1,j)-query(i-1,j-1)+arr[(i-1)*m+j];
			}
		}
		rep2(i,1,n,1){
			rep2(j,1,m,1){
				if(arr[(i-1)*m+j]!=0){
					arr[(i-1)*m+j]=1;
				}
			}
		}
		rep2(i,1,n,1){
			rep2(j,1,m,1){
				arr[(i-1)*m+j]=query(i,j-1)+query(i-1,j)-query(i-1,j-1)+arr[(i-1)*m+j];
			}
		}		
		int q;
		scanf("%d",&q);
		while(q--){
			int x1,y1,x2,y2;
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			int sum=query(x2,y2)-query(x1-1,y2)-query(x2,y1-1)+query(x1-1,y1-1);//覆盖面积
			int sum1=(x2-x1+1)*(y2-y1+1); //算上边上的点，所以加1
			if(sum==sum1) printf("YES\n");
			else printf("NO\n"); 
		}
	} 
    return 0;
}    
         

 

参考代码：https://blog.youkuaiyun.com/sdauguanweihong/article/details/89413462

 

二维数组前缀和：https://blog.youkuaiyun.com/qq_34990731/article/details/82807870