poj-3667 Hotel （线段树区间合并）

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 D_i (1 ≤ D_i ≤ N) and approach the front desk to check in. Each group i requests a set of D_i contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+D_i-1 if they are available or, if no contiguous set of rooms is available, politely suggests alternate lodging. Canmuu always chooses the value of rto be the smallest possible.

Visitors also depart the hotel from groups of contiguous rooms. Checkout i has the parameters X_i and D_i which specify the vacating of rooms X_i ..X_i +D_i-1 (1 ≤ X_i ≤ N-D_i+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 D_i (b) Three space-separated integers representing a check-out: 2, X_i, and D_i

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 61 31 31 31 32 5 51 6

Sample Output

14705

  【题意】N间空房子，共有M次操作，接下来M行，每行输入一个整数a，a=1时再输入一个数b，查询N间房子里面是否有b间连续的空房子，如果有返回最左边的空房间编号，如果a=2，接着输入两个数b,c，表示将[b,c]区间所有房间清空。典型的区间合并题，代码相对较长，还是有点小吃力的。

【AC】代码：

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<queue>#include<map>using namespace std;const int N=50000;int m,n;struct tree{    int l,r,lsum,rsum,ssum;    int ck;}node[N<<2];int max(int x,int y){    return x>y?x:y;}void pushup(int ans){    int ll = node[ans<<1].r - node[ans<<1].l+1;    int rr = node[ans<<1|1].r - node[ans<<1|1].l+1;    node[ans].lsum = node[ans<<1].lsum;    if(ll == node[ans<<1].lsum) node[ans].lsum+=node[ans<<1|1].lsum;    node[ans].rsum = node[ans<<1|1].rsum ;    if(rr == node[ans].rsum) node[ans].rsum+=node[ans<<1].rsum;    node[ans].ssum = max(max(node[ans<<1].ssum, node[ans<<1|1].ssum),node[ans<<1].rsum+node[ans<<1|1].lsum);}void pushdown(int ans){    if(node[ans].ck!=-1)    {        int ll = node[ans<<1].r- node[ans<<1].l+1;        int rr = node[ans<<1|1].r - node[ans<<1|1].l+1;        node[ans<<1].ck = node[ans<<1|1].ck = node[ans].ck;        node[ans].ck = -1;        node[ans<<1].rsum = node[ans<<1].lsum =node[ans<<1].ssum = node[ans<<1].ck?0:ll;        node[ans<<1|1].rsum = node[ans<<1|1].lsum = node[ans<<1|1].ssum = node[ans<<1|1].ck?0:rr;    }}void build_tree(int ans,int ll,int rr){    node[ans].l =ll, node[ans].r =rr, node[ans].ck =-1;    node[ans].lsum = node [ans].rsum =node[ans].ssum = rr - ll +1 ;    if(ll == rr )        return ;    int mid = (ll + rr )>>1 ;    build_tree(ans<<1,ll,mid);    build_tree(ans<<1|1,mid+1,rr);    pushup(ans);}void update(int ans,int ll,int rr,int value){    if(node[ans].l == ll&& node[ans].r == rr)    {        node[ans].ssum = node[ans].lsum = node[ans].rsum = value?0:(rr-ll+1);        node[ans].ck = value;        return ;    }    pushdown(ans);    int m = (node[ans].l + node[ans].r)>>1;    if(rr<=m)    {        update(ans<<1,ll,rr,value);    }    else if(ll>m)    {        update(ans<<1|1,ll,rr,value);    }    else    {        update(ans<<1,ll,m,value);        update(ans<<1|1,m+1,rr,value);    }    pushup(ans);}int query(int ans,int vis){    if(node[ans].l == node[ans].r) return node[ans].l;    pushdown(ans);    int mid= (node[ans].l+ node[ans].r)>>1;    if(node[ans<<1].ssum>=vis) return query(ans<<1,vis);    else if(node[ans<<1].rsum+node[ans<<1|1].lsum>=vis) return mid-node[ans<<1].rsum+1;    return query(ans<<1|1,vis);}int main(){    scanf("%d%d",&m,&n);    build_tree(1,1,m);    while(n--)    {        int x;        scanf("%d",&x);        if(x==1)        {            int w;            scanf("%d",&w);            if(node[1].ssum<w) puts("0");//如果根节点的最大连续空余房间数小于w，则直接输出0            else            {                int p=query(1,w);                printf("%d\n",p);                update(1,p,p+w-1,1);//区间更新              }        }        else        {            int a,b;            scanf("%d %d",&a,&b);            update(1,a,a+b-1,0); //区间更新        }    }    return 0;}