poj 3667 Hotel(线段树，区间合并)
Hotel
Time Limit: 3000MS
Memory Limit: 65536K
Total Submissions: 10858
Accepted: 4691
Description
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.
Input
* 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
Output
* 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.
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：告诉你有n个房间,一开始都是空房间,接下来m次操作，每一次操作，如果第一个数是1，Di就表示有人需要租Di间连续的房间，如果没有输出0，如果有则房间租出，
如果有满足条件的有种情况，则租最左边的。如果第一个数是2，则Xi、Di分别表示从Xi号房间到Xi+Di-1号房间的人退房。
：：这道题我是在参考hh神的代码慢慢敲出来的，现在有些理解
房间 1 2 3 4 5 6 7 8
（1租出） 1 1 0 1 1 1 1 1
lsum表示的是区间左端开始的连续房间个数，即lsum=2;
rsum表示的是区间右端开始的连续房间个数，即rsum=5;
msum表示中间部分连续房间的最大个数,即msum=5;
代码：
1: #include <iostream>
2: #include <cstdio>
3: #include <cstring>
4: #include <algorithm>
5: #include <queue>
6: using namespace std;
7: #define lson l,m,rt<<1
8: #define rson m+1,r,rt<<1|1
9: typedef long long ll;
10: const int N=55555;
11: int lsum[N<<2],rsum[N<<2],msum[N<<2];
12: int cover[N<<2];
13:
14: void Down(int rt,int m)
15: {
16: if(cover[rt]!=-1)
17: {
18: int ls=rt<<1,rs=ls|1;
19: cover[ls]=cover[rs]=cover[rt];
20: lsum[ls]=rsum[ls]=msum[ls]=cover[rt]?0:m-(m>>1);
21: lsum[rs]=rsum[rs]=msum[rs]=cover[rt]?0:(m>>1);
22: cover[rt]=-1;
23: }
24: }
25:
26: void Up(int rt,int m)
27: {
28: lsum[rt]=lsum[rt<<1];
29: rsum[rt]=rsum[rt<<1|1];
30: if(lsum[rt]==m-(m>>1)) lsum[rt]+=lsum[rt<<1|1];
31: if(rsum[rt]==(m>>1)) rsum[rt]+=rsum[rt<<1];
32: msum[rt]=max(rsum[rt<<1]+lsum[rt<<1|1],max(msum[rt<<1],msum[rt<<1|1]));
33: }
34:
35: void build(int l,int r,int rt)
36: {
37: lsum[rt]=rsum[rt]=msum[rt]=r-l+1;
38: cover[rt]=-1;
39: if(l==r) return ;
40: int m=(l+r)>>1;
41: build(lson);
42: build(rson);
43: }
44:
45: void update(int L,int R,int c,int l,int r,int rt)
46: {
47: if(L<=l&&R>=r)
48: {
49: lsum[rt]=rsum[rt]=msum[rt]=c?0:r-l+1;
50: cover[rt]=c;
51: return ;
52: }
53: Down(rt,r-l+1);
54: int m=(l+r)>>1;
55: if(L<=m) update(L,R,c,lson);
56: if(R>m) update(L,R,c,rson);
57: Up(rt,r-l+1);
58: }
59:
60: int query(int len,int l,int r,int rt)
61: {
62: if(l==r) return l;
63: Down(rt,r-l+1);
64: int m=(l+r)>>1;
65: if(msum[rt<<1]>=len) return query(len,lson);
66: else if(rsum[rt<<1]+lsum[rt<<1|1]>=len)
67: return m-rsum[rt<<1]+1;
68: return query(len,rson);
69: }
70:
71: int main()
72: {
73: // freopen("in.txt","r",stdin);
74: int n,m;
75: while(scanf("%d%d",&n,&m)>0)
76: {
77: int c,a,b;
78: build(1,n,1);
79: while(m--)
80: {
81: scanf("%d",&c);
82: if(c==1)
83: {
84: scanf("%d",&a);
85: if(msum[1]<a) puts("0");
86: else
87: {
88: int p=query(a,1,n,1);
89: printf("%d\n",p);
90: update(p,p+a-1,1,1,n,1);
91: }
92: }
93: else
94: {
95: scanf("%d%d",&a,&b);
96: update(a,a+b-1,0,1,n,1);
97: }
98: }
99: }
100: return 0;
101: }