Poj 1823 Hotel line segment tree. Pay attention to the lazy flag. If it is not marked, the drops will time out.

[Cpp]
The answer is 491 accepted, which is quite scary. Don't be scared. Actually, I am scared. Hey, it's also a tragedy to do this, however, when I look at the category, it is a medium question, so I am so angry that I am a third.
We need to use ml, mr, len on the online segment to indicate several consecutive vacancies on the left, several consecutive vacancies on the right, and several consecutive vacancies in the middle.
Pay attention to updating rules when merging
[Cpp]
A [I]. len = max (a [I * 2]. mr + a [I * 2 + 1]. ml, max (a [I * 2]. len, a [I * 2 + 1]. len); // update the len value.
If (a [I * 2 + 1]. ml = (a [I * 2 + 1]. r-a [I * 2 + 1]. l + 1) // If the ml of the child on the right is equal to the length of the Line Segment, that is, the child on the right is empty, then a [I]. ml is a [I * 2]. the entire length on the right side of mr +
A [I]. mr = a [I * 2]. mr + a [I * 2 + 1]. ml;
Else
A [I]. mr = a [I * 2 + 1]. mr; // if the right side of the child is empty, then a [I]. long mr is equal to a [I * 2 + 1]. mr
If (a [I * 2]. mr = (a [I * 2]. r-a [I * 2]. l + 1) // same as above
A [I]. ml = a [I * 2 + 1]. ml + a [I * 2]. mr;
Else
A [I]. ml = a [I * 2]. ml ;//

There is also a lazy flag. If it is not marked, it will time out.
[Cpp]
# Include <iostream>
# Include <cstdio>
Using namespace std;
# Define N 16005
Struct node {
Int l, r, ml, mr, mm, len, sign;
} A [N * 4];
Void build (int I, int left, int right ){
A [I]. l = left;
A [I]. r = right;
A [I]. len = a [I]. ml = a [I]. mr = right-left + 1;
A [I]. mm = 0;
A [I]. sign = 0; // indicates whether the edge is used.
If (left = right) return;
Int mid = (a [I]. l + a [I]. r)> 1;
Build (I * 2, left, mid );
Build (I * 2 + 1, mid + 1, right );
}
Void insert (int I, int left, int right, int sign ){
// Cout <a [I]. l <"" <a [I]. r <"****" <left <"" <right <endl;
If (a [I]. l> = left & a [I]. r <= right ){
A [I]. sign = 1;
If (sign = 1 ){
A [I]. len = a [I]. ml = a [I]. mr = 0;
} Else {
A [I]. len = a [I]. ml = a [I]. mr = (a [I]. r-a [I]. l + 1 );
}
Return;
}
If (a [I]. sign = 1 & a [I]. ml = 0 & a [I]. mr = 0 & a [I]. len = 0) {// lazy mark
A [I * 2 + 1]. ml = a [I * 2 + 1]. mr = a [I * 2 + 1]. len = 0;
A [I * 2]. ml = a [I * 2]. mr = a [I * 2]. len = 0;
A [I * 2 + 1]. sign = a [I * 2]. sign = 1;
A [I]. sign = 0;
}
If (a [I]. sign = 1 & a [I]. ml = (a [I]. r-a [I]. l + 1) & a [I]. mr = (a [I]. r-a [I]. l + 1) & a [I]. len = (a [I]. r-a [I]. l + 1) {// lazy mark
A [I * 2 + 1]. ml = a [I * 2 + 1]. mr = a [I * 2 + 1]. len = (a [I * 2 + 1]. r-a [I * 2 + 1]. l + 1 );
A [I * 2]. ml = a [I * 2]. mr = a [I * 2]. len = (a [I * 2]. r-a [I * 2]. l + 1 );
A [I * 2 + 1]. sign = a [I * 2]. sign = 1;
A [I]. sign = 0;
}
Int mid = (a [I]. l + a [I]. r)> 1;
If (right <= mid) insert (I * 2, left, right, sign );
Else if (left> mid) insert (I * 2 + 1, left, right, sign );
Else {
Insert (I * 2, left, mid, sign );
Insert (I * 2 + 1, mid + 1, right, sign );
}
A [I]. len = max (a [I * 2]. mr + a [I * 2 + 1]. ml, max (a [I * 2]. len, a [I * 2 + 1]. len ));
If (a [I * 2 + 1]. ml = (a [I * 2 + 1]. r-a [I * 2 + 1]. l + 1 ))
A [I]. mr = a [I * 2]. mr + a [I * 2 + 1]. ml;
Else
A [I]. mr = a [I * 2 + 1]. mr;
If (a [I * 2]. mr = (a [I * 2]. r-a [I * 2]. l + 1 ))
A [I]. ml = a [I * 2 + 1]. ml + a [I * 2]. mr;
Else
A [I]. ml = a [I * 2]. ml;
// Cout <"l =" <a [I]. l <"r =" <a [I]. r <"" <a [I]. ml <"" <a [I]. len <"" <a [I]. mr <endl;
}
 
Int main (){
Int n, p, m, x, y;
While (~ Scanf ("% d", & n, & p )){
Build (1, 1, n );
While (p --){
Scanf ("% d", & m );
If (m = 1 ){
Scanf ("% d", & x, & y );
// Cout <"x + Y-1 =" <x + Y-1 <endl;
Insert (1, x, x + Y-1, 1 );
} Else if (m = 2 ){
Scanf ("% d", & x, & y );
Insert (1, x, x + Y-1, 0 );
}
Else
Printf ("% d \ n", max (a [1]. ml, max (a [1]. len, a [1]. mr )));
}
 
}
}

Author: youngyangyang04