Hotel ----线段树并查集，区间合并

Hotel

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 11223		Accepted: 4855

DescriptionThe 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 DiOutput* 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

Source题意：有N个房间，M次操作。1 a表示找到连续的长度为a的空房间，如果有多解，优先左边的，即表示入住。2 b len把起点为b长度的len的房间清空，即退房。解题报告：对于道题来说重点是分析怎么是连续长度为a的空房间，如果查询存在这样的空房间的话，就将[a,a+b-1]进行清空。代码：#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cctype>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int MAXN = 55555;int lsum[MAXN<<2],rsum[MAXN<<2],msum[MAXN<<2];int cover[MAXN<<2];void PushDown(int rt,int m){if(cover[rt]!=-1){cover[rt<<1]=cover[rt<<1|1]=cover[rt];msum[rt<<1]=lsum[rt<<1]=rsum[rt<<1]=cover[rt] ? 0 : m-(m>>1);msum[rt<<1|1]=lsum[rt<<1|1]=rsum[rt<<1|1]=cover[rt] ? 0 : (m>>1);cover[rt]=-1;}}void PushUp(int rt,int m){lsum[rt]=lsum[rt<<1];rsum[rt]=rsum[rt<<1|1];if(lsum[rt]==m-(m>>1))lsum[rt]+=lsum[rt<<1|1];if(rsum[rt]==(m>>1))rsum[rt]+=rsum[rt<<1];msum[rt]+=max(lsum[rt<<1|1]+rsum[rt<<1],max(msum[rt<<1],msum[rt<<1|1]));}void build(int l,int r,int rt){msum[rt]=lsum[rt]=rsum[rt]=r-l+1;cover[rt]=-1;if(l==r){return;}int m=(l+r)>>1;build(lson);build(rson);}void update(int L,int R,int c,int l,int r,int rt){if(L<=l && r<=R){msum[rt]=lsum[rt]=rsum[rt]=c ? 0 : r-l+1;cover[rt]=c;return;}PushDown(rt,r-l+1);int m=(l+r)>>1;if(L<=m){update(L,R,c,lson);}if(m<R){update(L,R,c,rson);}PushUp(rt,r-l+1);}int query(int w,int l,int r,int rt){if(l==r){return 1;}PushDown(rt,r-l+1);int m=(l+r)>>1;if(msum[rt<<1]>=w)return query(w,lson);else if(rsum[rt<<1]+lsum[rt<<1|1]>=w){return m-rsum[rt<<1]+1;}return query(w,rson);}int main(){int n,m;scanf("%d%d",&n,&m);build(1,n,1);while(m--){int op,a,b;scanf("%d",&op);if(op==1){scanf("%d",&a);if(msum[1]<a)puts("0");else{int p=query(a,1,n,1);printf("%d\n",p);update(p,p+a-1,1,1,n,1);}}else{scanf("%d%d",&a,&b);update(a,a+b-1,0,1,n,1);}}return 0;}