Add Tutorial for LOJ-1266: Points in Rectangle (en)

1266/en.md

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+# LightOj 1266 - Points in Rectangle
+### Tag
+Data Structure, Binary Indexed Tree
+### Quick links for prerequisites
+Binary Indexed Tree:  
+- https://www.geeksforgeeks.org/binary-indexed-tree-or-fenwick-tree-2
+- https://cp-algorithms.com/data_structures/fenwick.html
+
+
+### Solution 
+- We will maintain a 2-D BIT to solve this problem. For query 0 x y, we will call update function each time we got a new point (x,y). For the second query, we will call query function with parameter (x,y). It will return  the sum of the rectangle that's lower left co-ordinate is (1,1) and upper right co-ordinate is (x,y). Let's take a look first testcase: <br>
+1   <br>
+4   <br> 
+0 1 1   <br>
+0 2 6   <br>
+1 1 1 6 6   <br>
+1 2 2 5 5   <br>
+-  We got two points (1,1) and (2,6), let's plot these points <br>
+![first](./first.PNG)
+- In the first query , we have to find how many point are inside (1,1) and (6,6) . We will call query funtion with parameter (6,6) , it will return the number of point. So the ans is 2. <br>
+- Let's come to the second query. Here we have to find the number of point inside (2,2) and (5,5). We will call query function with parameter (5,5). But it will return the number of point inside (1,1) and (5,5). The area is ploted bellow in red-  
+  ![second](./second.PNG)
+- We took some extra area. Because we have to find from (2,2) to (5,5) not (1,1) to (5,5). However, the final equation of ans is - <br>
+Ans=**A-B-C+D** Where <br>  A=query(x2,y2) <br> B=query(x1-1,y2-1) <br> C=query(x2-1,y1-1) <br> D=query(x1-1,y1-1) <br>
+let's clarify this equation. First look at the area ploted bellow of **A,B,C,D**. 
+ ![second](./third.PNG)
+ ![second](./fourth.PNG)
+- As D is inside of both B and C. So at the time of substracting B and C , D is subtracted twice . so D is added. Now after calculating 
+we got our area where no point is located. So ans of the second query is 0. Hope you understood the solution.
+
+### Code
+
+#### C++
+```cpp
+#include<bits/stdc++.h>
+using namespace std;
+long long int BIT[1005][1005];
+bool vis[1005][1005];
+
+void update(int x, int y , int val)
+{
+    while(x<=1001)
+    {
+        int y1=y;
+        while(y1<=1001)
+        {
+            BIT[x][y1]+=val;
+            y1+=(y1&-y1);
+        }
+        x+=(x&-x);
+    }
+}
+long long int query(int x, int y)
+{
+    long long int sum=0;
+    while(x>0)
+    {
+        int y1=y;
+        while(y1>0)
+        {
+            sum+=BIT[x][y1];
+            y1-=(y1&-y1);
+        }
+        x-=(x&-x);
+    }
+    return sum;
+}
+void solve()
+{
+
+   memset(BIT,0,sizeof(BIT)); //initialize with zero 
+   memset(vis,false,sizeof(vis)); // initalize with zero
+   long long int q,a,b,c,d,x1,y1,x2,y2;
+   cin>>q;
+   while(q--)
+   {
+      cin>>a;
+      if(a==0)
+      {
+         cin>>a>>b;
+         a++,b++;
+         if(!vis[a][b])
+         {
+            vis[a][b]=true; 
+            update(a,b,1); //update the index 
+         }
+      }
+      else
+      {
+         cin>>x1>>y1>>x2>>y2;
+         x1++,x2++,y1++,y2++;
+         long long int ans=query(x2,y2)-query(x2,y1-1)-query(x1-1,y2)+query(x1-1,y1-1); /* find total point inside the given rectangle */
+         cout<<ans<<endl;
+      }
+   }
+   return;
+}
+int32_t main()
+{
+  int tt;
+  tt = 1;
+  cin>>tt;
+  for(int i=1;i<=tt;i++)
+  {
+     cout<<"Case "<<i<<":"<<endl;
+     solve();
+  }
+  return 0;
+}
+```

1266/solve.txt

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+# LightOj 1266 - Points in Rectangle
+### Tag
+Data Structure, Binary Indexed Tree
+### Quick links for prerequisites
+Binary Indexed Tree:  
+- https://www.geeksforgeeks.org/binary-indexed-tree-or-fenwick-tree-2
+- https://cp-algorithms.com/data_structures/fenwick.html
+
+
+### Solution 
+- We will maintain a 2-D BIT to solve this problem. For query 0 x y, we will call update function each time we got a new point (x,y). For the second query, we will call query function with parameter (x,y). It will return  the sum of the rectangle that's lower left co-ordinate is (1,1) and upper right co-ordinate is (x,y). Let's take a look first testcase: <br>
+1   <br>
+4   <br>
+0 1 1   <br>
+0 2 6   <br>
+1 1 1 6 6   <br>
+1 2 2 5 5   <br>
+-  We got two points (1,1) and (2,6), let's plot these points <br>
+![first](https://github.com/Rabby033/problem-tutorials/assets/52863153/5bdf2a95-34db-4f0e-88df-8a1c0cd8aa8f)
+- In the first query , we have to find how many point are inside (1,1) and (6,6) . We will call query funtion with parameter (6,6) , it will return the number of point. So the ans is 2. <br>
+- Let's come to the second query. Here we have to find the number of point inside (2,2) and (5,5). We will call query function with parameter (5,5). But it will return the number of point inside (1,1) and (5,5). The area is ploted bellow in red-  
+  ![second](https://github.com/Rabby033/problem-tutorials/assets/52863153/58e666ac-42f8-41a9-8381-d129e94b23d2)
+- We took some extra area. Because we have to find from (2,2) to (5,5) not (1,1) to (5,5). However, the final equation of ans is - <br>
+Ans=**A-B-C+D** Where <br>  A=query(x2,y2) <br> B=query(x1-1,y2-1) <br> C=query(x2-1,y1-1) <br> D=query(x1-1,y1-1) <br>
+let's clarify this equation. First look at the area ploted bellow of **A,B,C,D**. 
+![third](https://github.com/Rabby033/problem-tutorials/assets/52863153/ddeec284-bedd-428f-aa94-09484b3f668e)
+![fourth](https://github.com/Rabby033/problem-tutorials/assets/52863153/781a3c06-2646-46f0-a643-f58d175450a1)
+- As D is inside of both B and C. So at the time of substracting B and C , D is subtracted twice . so D is added. Now after calculating 
+we got our area where no point is located. So ans of the second query is 0. Hope you understood the solution.
+
+### Code
+
+#### C++
+```cpp
+#include<bits/stdc++.h>
+using namespace std;
+long long int BIT[1005][1005];
+bool vis[1005][1005];
+
+void update(int x, int y , int val)
+{
+    while(x<=1001)
+    {
+        int y1=y;
+        while(y1<=1001)
+        {
+            BIT[x][y1]+=val;
+            y1+=(y1&-y1);
+        }
+        x+=(x&-x);
+    }
+}
+long long int query(int x, int y)
+{
+    long long int sum=0;
+    while(x>0)
+    {
+        int y1=y;
+        while(y1>0)
+        {
+            sum+=BIT[x][y1];
+            y1-=(y1&-y1);
+        }
+        x-=(x&-x);
+    }
+    return sum;
+}
+void solve()
+{
+
+   memset(BIT,0,sizeof(BIT)); //initialize with zero 
+   memset(vis,false,sizeof(vis)); // initalize with zero
+   long long int q,a,b,c,d,x1,y1,x2,y2;
+   cin>>q;
+   while(q--)
+   {
+      cin>>a;
+      if(a==0)
+      {
+         cin>>a>>b;
+         a++,b++;
+         if(!vis[a][b])
+         {
+            vis[a][b]=true; 
+            update(a,b,1); //update the index 
+         }
+      }
+      else
+      {
+         cin>>x1>>y1>>x2>>y2;
+         x1++,x2++,y1++,y2++;
+         long long int ans=query(x2,y2)-query(x2,y1-1)-query(x1-1,y2)+query(x1-1,y1-1); /* find total point inside the given rectengle */
+         cout<<ans<<endl;
+      }
+   }
+   return;
+}
+int32_t main()
+{
+  int tt;
+  tt = 1;
+  cin>>tt;
+  for(int i=1;i<=tt;i++)
+  {
+     cout<<"Case "<<i<<":"<<endl;
+     solve();
+  }
+  return 0;
+}
+```