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：
在一个面积不超过n*m的矩形上，有p个矩形A，问之后的q个矩形B能否被之前的A全部覆盖（每个B的点都在至少一个A中）
分析：
官方题解：
由于n*m,p,q的范围过大，于是考虑O(n*m+p+q)的做法。
对于A类矩形(x1,y1,x2,y2)，我们只需要在(x1,y1),(x2+1,y2+1)处+1，在（x1,y2+1）(x2+1,y1)处-1
之后对整个面积求一个前缀和。则大于0的地方就是被A类矩形覆盖的点。
把值大于0的地方变成1，再一次求一次前缀和，处理好后即可在O(1)的时间算出一个矩形内被覆盖的点的数量。
哎，自己绕到二维线段树上去了，该大胆的试试的。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf=1e18;
const int N = 20000050;
//ll sum[N]{N];
ll sum[N];//二维变一维
int n,m;
/*//二维前缀和
int init() {
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
sum[i][j] = sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1] + a[i][j];
}
}
}
int get(int x1, int y1, int x2, int y2) {
return sum[x1][y1] - sum[x1][y2 - 1] - sum[x2 - 1][y1] + sum[x2 - 1][y2 - 1];
}*/
//二维差分
//将矩阵{（x1,y1)-(x2,y2)}的数+num
void add_1(int i,int j,int val)
{
if(i>n||j>m)
return ;
sum[(i-1)*m+j]+=val;
}
void add(int x1, int y1, int x2, int y2, int num)
{
//sum[x1][y1] += num;
//sum[x1][y2 + 1] -= num;
//sum[x2 + 1][y1] -= num;
//sum[x2 + 1][y2 + 1] += num;
//二维转一维
add_1(x1,y1,1);
add_1(x1,y2+1,-1);
add_1(x2+1,y1,-1);
add_1(x2+1,y2+1,1);
}
int query(int i,int j)
{
if(i==0||j==0)
return 0;
return sum[(i-1)*m+j];
}
//获取二维前缀和
void get()
{
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= m; j++)
{
//sum[i][j] += sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1];
sum[(i-1)*m+j]=query(i,j-1)+query(i-1,j)-query(i-1,j-1)+sum[(i-1)*m+j];//二维转一维
}
}
}
int main()
{
int x1,x2,y1,y2,p;
while(~scanf("%d%d%d",&n,&m,&p))
{
memset(sum,0,sizeof(sum));
for(int i=1; i<=p; i++)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
add(x1,y1,x2,y2,1);
}
get();
for(int i = 1 ; i <=n ; i++)
for(int j = 1 ; j <=m ; j++)
if(sum[(i-1)*m+j]>0)
sum[(i-1)*m+j]=1;
get();
int q;
scanf("%d",&q);
while(q--)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
int ans=query(x1-1,y1-1)-query(x1-1,y2)-query(x2,y1-1)+query(x2,y2);
printf("%s\n",(ans==(x2-x1+1)*(y2-y1+1))?"YES":"NO");
}
}
return 0;
}
