酒店
题目描述
The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny shores of Lake Superior. Bessie, ever the competent travel agent, has named the Bullmoose Hotel on famed Cumberland Street as their vacation residence. This immense hotel has N (1 ≤ N ≤ 50,000) rooms all located on the same side of an extremely long hallway (all the better to see the lake, of course).
The cows and other visitors arrive in groups of size Di (1 ≤ Di ≤ N) and approach the front desk to check in. Each group i requests a set of Di contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+Di-1 if they are available or, if no contiguous set of rooms is available, politely suggests alternate lodging. Canmuu always chooses the value of r to be the smallest possible.
Visitors also depart the hotel from groups of contiguous rooms. Checkout i has the parameters Xi and Di which specify the vacating of rooms Xi ..Xi +Di-1 (1 ≤ Xi ≤ N-Di+1). Some (or all) of those rooms might be empty before the checkout.
Your job is to assist Canmuu by processing M (1 ≤ M < 50,000) checkin/checkout requests. The hotel is initially unoccupied.
参考样例，第一行输入n，m ，n代表有n个房间，编号为1---n，开始都为空房，m表示以下有m行操作，以下 每行先输入一个数 i ，表示一种操作：
若i为1，表示查询房间，再输入一个数x，表示在1--n 房间中找到长度为x的连续空房，输出连续x个房间中左端的房间号，尽量让这个房间号最小，若找不到长度为x的连续空房，输出0。
若i为2，表示退房，再输入两个数 x，y 代表 房间号 x---x+y-1 退房，即让房间为空。
输入输出格式
输入格式：
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Line i+1 contains request expressed as one of two possible formats: (a) Two space separated integers representing a check-in request: 1 and Di (b) Three space-separated integers representing a check-out: 2, Xi, and Di
输出格式：
* Lines 1.....: For each check-in request, output a single line with a single integer r, the first room in the contiguous sequence of rooms to be occupied. If the request cannot be satisfied, output 0.
输入输出样例
1
4
7
0
5
分析：
可以说这一类题目就是检验自己线段树是否真正理解透彻的测验。
对于这题，我们需要维护四个信息，$sum$表示这一段最大的连续空房间数，$sl$表示这一段从左数的最大连续空房间数，$sr$表示这一段从右数的最大连续空房间数，$sign$表示这段区间是否被修改。
在维护信息时，$pushup$操作如下，详细解释在代码中：
inline void pushup(int rt) { if( t[ls].sum==t[ls].len ) t[rt].sl=t[ls].sum+t[rs].sl;//如果左子树全为空的情况 else t[rt].sl=t[ls].sl; if( t[rs].sum==t[rs].len ) t[rt].sr=t[rs].sum+t[ls].sr;//右子树全为空的情况 else t[rt].sr=t[rs].sr; t[rt].sum=max(max(t[ls].sum,t[rs].sum),t[ls].sr+t[rs].sl); //求最大连续空房间数。 }
$pushdown$操作如下：
inline void pushdown(int rt) { if( t[rt].sign==0 ) return; if( t[rt].sign==1 ) {//如果sign为1，则房间已经被占用了 t[ls].sum=t[ls].sl=t[ls].sr=0; t[rs].sum=t[rs].sl=t[rs].sr=0; t[ls].sign=t[rs].sign=1; } if( t[rt].sign==-1 ) {//如果sign为-1，则房间要空出来 t[ls].sum=t[ls].sl=t[ls].sr=t[ls].len; t[rs].sum=t[rs].sl=t[rs].sr=t[rs].len; t[ls].sign=t[rs].sign=-1; } t[rt].sign=0; }
再结合全部代码应该就很好理解了。
Code：
//It is made by HolseLee on 5th Nov 2018 //Luogu.org P2894 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ls (rt<<1) #define rs (rt<<1|1) using namespace std; const int N=1e5+7; int n,m; struct Segment { int sum,sl,sr,len,sign; }t[N<<2]; inline int read() { char ch=getchar(); int x=0; bool flag=false; while( ch<'0' || ch>'9' ) { if( ch=='-' ) flag=true; ch=getchar(); } while( ch>='0' && ch<='9' ) { x=x*10+ch-'0'; ch=getchar(); } return flag ? -x : x; } inline void pushup(int rt) { if( t[ls].sum==t[ls].len ) t[rt].sl=t[ls].sum+t[rs].sl; else t[rt].sl=t[ls].sl; if( t[rs].sum==t[rs].len ) t[rt].sr=t[rs].sum+t[ls].sr; else t[rt].sr=t[rs].sr; t[rt].sum=max(max(t[ls].sum,t[rs].sum),t[ls].sr+t[rs].sl); } inline void pushdown(int rt) { if( t[rt].sign==0 ) return; if( t[rt].sign==1 ) { t[ls].sum=t[ls].sl=t[ls].sr=0; t[rs].sum=t[rs].sl=t[rs].sr=0; t[ls].sign=t[rs].sign=1; } if( t[rt].sign==-1 ) { t[ls].sum=t[ls].sl=t[ls].sr=t[ls].len; t[rs].sum=t[rs].sl=t[rs].sr=t[rs].len; t[ls].sign=t[rs].sign=-1; } t[rt].sign=0; } void build(int l,int r,int rt) { t[rt].sign=0; t[rt].sum=t[rt].sl=t[rt].sr=t[rt].len=r-l+1; if( l==r ) return; int mid=(l+r)>>1; build(l,mid,ls); build(mid+1,r,rs); } void update(int l,int r,int rt,int L,int R,int C) { pushdown(rt); if( L<=l && r<=R ) { if( C==1 ) t[rt].sum=t[rt].sl=t[rt].sr=0; else t[rt].sum=t[rt].sl=t[rt].sr=t[rt].len; t[rt].sign=C; return; } int mid=(l+r)>>1; if( L<=mid ) update(l,mid,ls,L,R,C); if( R>mid ) update(mid+1,r,rs,L,R,C); pushup(rt); } int query(int l,int r,int rt,int C) { pushdown(rt); if( l==r ) return l; int mid=(l+r)>>1; if( t[ls].sum>=C ) return query(l,mid,ls,C); else if( t[ls].sr+t[rs].sl>=C ) return (mid-t[ls].sr+1); else return query(mid+1,r,rs,C); } int main() { n=read(), m=read(); int opt,x,y; build(1,n,1); for(int i=1; i<=m; ++i) { opt=read(); if( opt==1 ) { x=read(); if( t[1].sum<x ) { puts("0"); continue; } printf("%d\n",y=query(1,n,1,x)); update(1,n,1,y,y+x-1,1); } else { x=read(), y=read(); update(1,n,1,x,x+y-1,-1); } } return 0; }