POJ 3667 Hotel 【线段树 区间合并 + Lazy-tag】
Hotel
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Time Limit: 3000MS |
Memory Limit: 65536K |
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Description
The cows are journeying north to ThunderBay in Canada to gain cultural enrichment and enjoy a vacation on the sunnyshores of Lake Superior. Bessie, ever the competent travel agent, has named theBullmoose Hotel on famed Cumberland Street as their vacation residence. Thisimmense hotel has N (1 ≤ N ≤ 50,000) roomsall located on the same side of an extremely long hallway (all the better tosee the lake, of course).
The cows and other visitors arrive ingroups of size Di (1 ≤ Di ≤N) and approach the front desk to check in. Each group i requestsa set of Di contiguous rooms from Canmuu, the moosestaffing the counter. He assigns them some set of consecutive room numbers r..r+Di-1if they are available or, if no contiguous set of rooms is available, politely suggestsalternate lodging. Canmuu always chooses the value of r to bethe smallest possible.
Visitors also depart the hotel from groupsof contiguous rooms. Checkout i has the parameters Xi and Di whichspecify 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 isinitially unoccupied.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Line i+1 contains request expressed as oneof two possible formats: (a) Two space separated integers representing acheck-in request: 1 and Di (b) Threespace-separated integers representing a check-out: 2, Xi,and Di
Output
* Lines 1.....: For each check-in request,output a single line with a single integer r, the first room in thecontiguous sequence of rooms to be occupied. If the request cannot besatisfied, output 0.
Sample Input
10 6
1 3
1 3
1 3
1 3
2 5 5
1 6
Sample Output
1
4
7
0
5
题意：
有一个Hotel。有N个连续的房间，開始的时候没有人住。以下有M次‘1’或‘2’操作。
操作“1 D”表示有D个人进来选房间，他们必须是住在连号的房间，假设不满足这个条件，输出“0”，否则，输出他们这几个人中房间号码最小的一个；操作“2 X D”表示Hotel从房间号X開始D个房间被清空。
分析：
事实上操作‘1’能够看成是两步，首先，我得找到[1,N]区间中是否包括有长度为D连续的未被占用的房间的区间，而且返回区间的首元素。
我的方法是线段树区间合并。因为在我另外一篇博客《hdu 1540/POJ 2892 Tunnel Warfare 【线段树区间合并】》中有讲到类似的做法，我这里不累赘太多，仅仅只是这个题目相对复杂一点，主要是多了一个Lazy-tag的思想。
线段树题目的类型大致能够分为四种：单点更新、成段增减或更新、区间合并和扫描线
成段更新和区间合并都须要用到Lazy-tag思想。
扫描线就是求矩形面积和周长的题目，须要用到离散化。
本篇解说区间合并，区间合并肯定是从子节点向上才干用着合并，这类题目都是求最长连续区间的。主要在PushUp的时候须要对左右儿子的区间进行合并。
设对于每个节点rt，它所管辖的范围是 [l,r],我这里用相似的对每个节点用ln，rn，mn 分别来表示在[l,r] 中以l 開始的连续未被占用的区间长度。以 r 结尾的连续且未被占用的区间长度，以及[l,r]这个区间最大的连续未被占用的区间长度。注意。我这里都是指未被占用哦~ 【详细区间合并的解说。请看《hdu 1540/POJ 2892 Tunnel Warfare 【线段树区间合并】》中的解说~~~
这个题目，须要对区间进行更新，区间更新 <--> 延迟更新（Lazy-tag）；那么我这里的tag数组就是用来延迟更新的，tag[rt]= -1 表示清除标记时的状态。 tag[rt] =0 表示将区间所有占用时的状态。tag[rt] = 1 表示区间为空的状态，一般来说，更新的状态有i种。就相应着tag[rt] 的值就有i+1 种，由于当中一种是用来清除标记的~ O(∩_∩)O哈哈~
/****************************>>>>HEADFILES<<<<****************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <sstream>
#include <algorithm>
using namespace std;
/****************************>>>>>DEFINE<<<<<*****************************/
#define fst first
#define snd second
#define root 1,N,1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PB(a) push_back(a)
#define MP(a,b) make_pair(a,b)
#define CASE(T) for(scanf("%d",&T);T--;)
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
//#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef __int64 LL;
const int INF = 0x3f3f3f3f;
/****************************>>>>SEPARATOR<<<<****************************/
const int maxn = 50000 + 5;
int N, M;
struct Node
{
int ln, rn, mn;
} segtree[maxn << 2];// Have not been occupied
int tag[maxn << 2];//-1 without tag 0 FULL 1 Empty
inline void PushUp(const int& rt, const int& l, const int& r)
{
segtree[rt].ln = segtree[rt << 1].ln;
segtree[rt].rn = segtree[rt << 1 | 1].rn;
segtree[rt].mn = max(segtree[rt << 1].rn + segtree[rt << 1 | 1].ln,
max(segtree[rt << 1].mn, segtree[rt << 1 | 1].mn));
int mid = (l + r) >> 1;
if(segtree[rt << 1].mn == mid - l + 1)
segtree[rt].ln += segtree[rt << 1 | 1].ln;
if(segtree[rt << 1 | 1].mn == r - (mid + 1) + 1)
segtree[rt].rn += segtree[rt << 1].rn;
}
inline void PushDown(const int& rt, const int& l, const int& r)
{
if(tag[rt] != -1)
{
tag[rt << 1] = tag[rt << 1 | 1] = tag[rt];
int mid = (l + r) >> 1;
segtree[rt << 1].ln = segtree[rt << 1].rn = segtree[rt << 1].mn = tag[rt] * (mid - l + 1);
segtree[rt << 1 | 1].ln = segtree[rt << 1 | 1].rn = segtree[rt << 1 | 1].mn = tag[rt] * (r - (mid + 1) + 1);
tag[rt] = -1;
}
}
void Build(int l, int r, int rt)
{
segtree[rt].ln = segtree[rt].rn = segtree[rt].mn = r - l + 1;
tag[rt] = 1;
if(l == r) return ;
int mid = (l + r) >> 1;
Build(lson);
Build(rson);
}
void Update(int L, int R, const bool& isClear, int l, int r, int rt)
{
if(L <= l && r <= R)
{
if(isClear)
{
segtree[rt].ln = segtree[rt].rn = segtree[rt].mn = r - l + 1;
tag[rt] = 1;
}
else
{
segtree[rt].ln = segtree[rt].rn = segtree[rt].mn = 0;
tag[rt] = 0;
}
return;
}
PushDown(rt, l, r);
int mid = (l + r) >> 1;
if(L <= mid)
Update(L, R, isClear, lson);
if(mid < R)
Update(L, R, isClear, rson);
PushUp(rt, l, r);
}
int Query(const int& Len, int l, int r, int rt)
{
if(segtree[rt].ln >= Len) return l;
int mid = (l + r) >> 1;
if(segtree[rt << 1].mn >= Len)
return Query(Len, lson);
else if(segtree[rt << 1].rn + segtree[rt << 1 | 1].ln >= Len)
return mid - segtree[rt << 1].rn + 1;
else
return Query(Len, rson);
}
int Op, X, D, ans;
int main()
{
// FIN;
while(~scanf("%d %d", &N, &M))
{
Build(root);
while(M--)
{
scanf("%d", &Op);
if(Op & 1)
{
scanf("%d", &D);
if(segtree[1].mn < D)
{
printf("0\n");
continue;
}
ans = Query(D, root);
printf("%d\n", ans);
Update(ans, ans + D - 1, false, root);
}
else
{
scanf("%d %d", &X, &D);
Update(X, X + D - 1, true, root);
}
}
}
}
