POJ 1823 Hotel line segment tree + lazy label

Some rooms have three operations: one is continuous room occupancy, and the other is continuous room vacancy, the third is to ask how long the longest consecutive room is.
Idea: it is obviously the question of the Line Segment tree. lazy thought is used in the middle. A difficult question, which requires careful consideration. The key to this question is that after lazy is used to update down, the information of the parent node needs to be changed based on the information of the child node. That is to say, you also need to update from bottom up.
Code:
[Cpp]
# Include <iostream>
# Include <cstdio>
# Include <string. h>
# Include <algorithm>
Using namespace std;
 
Const int n= 16050;
Struct tree
{
Int lp, rp, llen, rlen, mlen, flag;
Int getmid ()
{
Return (lp + rp)/2;
}
} Tt [N * 4];
Void built_tree (int lp, int rp, int pos)
{
Tt [pos]. lp = lp;
Tt [pos]. rp = rp;
Tt [pos]. flag = 0;
If (lp = rp)
{
Tt [pos]. llen = tt [pos]. rlen = tt [pos]. mlen = 1;
Return;
}
Int mmid = tt [pos]. getmid ();
Built_tree (lp, mmid, pos * 2 );
Built_tree (mmid + 1, rp, pos * 2 + 1 );
Tt [pos]. llen = tt [pos]. rlen = tt [pos]. mlen = tt [pos * 2]. mlen + tt [pos * 2 + 1]. mlen;
}
Int max (int a, int B)
{
Return a> B? A: B;
}
Void update (int pos)
{
If (! Tt [pos * 2]. flag)
Tt [pos]. llen = tt [pos * 2]. mlen + tt [pos * 2 + 1]. llen;
Else
Tt [pos]. llen = tt [pos * 2]. llen;
If (! Tt [pos * 2 + 1]. flag)
Tt [pos]. rlen = tt [pos * 2]. rlen + tt [pos * 2 + 1]. mlen;
Else
Tt [pos]. rlen = tt [pos * 2 + 1]. rlen;
Int max1 = max (tt [pos]. llen, tt [pos]. rlen );
Int max2 = tt [pos * 2]. rlen + tt [pos * 2 + 1]. llen;
Int max3 = max (tt [pos * 2]. mlen, tt [pos * 2 + 1]. mlen );
Tt [pos]. mlen = max (max1, max (max2, max3 ));
If (tt [pos * 2]. flag = tt [pos * 2 + 1]. flag)
Tt [pos]. flag = tt [pos * 2]. flag;
}
Void insert (int lp, int rp, int pos)
{
If (tt [pos]. lp = lp & tt [pos]. rp = rp)
{
Tt [pos]. flag = 2;
Tt [pos]. llen = tt [pos]. rlen = tt [pos]. mlen = 0;
Return;
}
If (! Tt [pos]. flag)
{
Tt [pos]. flag = 1;
Tt [pos * 2]. flag = tt [pos * 2 + 1]. flag = 0;
Tt [pos * 2]. llen = tt [pos * 2]. rlen = tt [pos * 2]. mlen = tt [pos * 2]. rp-tt [pos * 2]. lp + 1;
Tt [pos * 2 + 1]. llen = tt [pos * 2 + 1]. rlen = tt [pos * 2 + 1]. mlen = tt [pos * 2 + 1]. rp-tt [pos * 2 + 1]. lp + 1;
}
Int mmid = tt [pos]. getmid ();
If (rp <= mmid)
{
Insert (lp, rp, pos * 2 );
}
Else if (lp> mmid)
{
Insert (lp, rp, pos * 2 + 1 );
}
Else
{
Insert (lp, mmid, pos * 2 );
Insert (mmid + 1, rp, pos * 2 + 1 );
}
Update (pos );
}
Void rem (int lp, int rp, int pos)
{
If (tt [pos]. lp = lp & tt [pos]. rp = rp)
{
Tt [pos]. flag = 0;
Tt [pos]. llen = tt [pos]. rlen = tt [pos]. mlen = tt [pos]. rp-tt [pos]. lp + 1;
Return;
}
If (tt [pos]. flag = 2)
{
Tt [pos]. flag = 1;
Tt [pos * 2]. flag = tt [pos * 2 + 1]. flag = 2;
Tt [pos * 2]. llen = tt [pos * 2]. rlen = tt [pos * 2]. mlen = 0;
Tt [pos * 2 + 1]. llen = tt [pos * 2 + 1]. rlen = tt [pos * 2 + 1]. mlen = 0;
}
Int mmid = tt [pos]. getmid ();
If (rp <= mmid)
Rem (lp, rp, pos * 2 );
Else if (lp> mmid)
Rem (lp, rp, pos * 2 + 1 );
Else
{
Rem (lp, mmid, pos * 2 );
Rem (mmid + 1, rp, pos * 2 + 1 );
}
Update (pos );
}
Int main ()
{
// Freopen ("1.txt"," r ", stdin );
Int n, m;
While (scanf ("% d", & n, & m )! = EOF)
{
Int id, x, y;
Built_tree (1, n, 1 );
While (m --)
{
Scanf ("% d", & id );
If (id = 3)
Printf ("% d \ n", tt [1]. mlen );
Else if (id = 1)
{
Scanf ("% d", & x, & y );
Insert (x, x + Y-1, 1 );
}
Else
{
Scanf ("% d", & x, & y );
Rem (x, x + Y-1, 1 );
}
}
}
Return 0;
}