Hotel poj 3667

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Hotel

 

 　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 18020		Accepted: 7823

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 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 r to 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 6
1 3
1 3
1 3
1 3
2 5 5
1 6

Sample Output

1
4
7
0
5

Source

USACO 2008 February Gold

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题意 ：

　　两种操作

　　　　1)  选一段 长度为 D 的可用区间 ，并占用它

　　　　2)  从 X 开始 长度为 D 的区间 标记为可用

　　开始时均为可用

　　区间范围 1--n

　　询问次数 m

　　1 ≤  n  ,m < 50,000

思路：

　　考虑区间合并线段树，带lazy标记

　　

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <functional>

#define lson l,mid,pos<<1
#define rson mid+1,r,pos<<1|1

using namespace std;

const int maxn = +;
int lazy[maxn<<];
int pre[maxn<<];
int suf[maxn<<];
int now[maxn<<];
int a[maxn];
void push_down(int l,int r,int pos)
{
    if (l==r){
        a[l]=lazy[pos]-;
        pre[pos]=suf[pos]=now[pos]=(a[l]==)*;
        lazy[pos]=;
        return ;
    }
    int mid=(l+r)>>;
    int lpos=pos<<,rpos=pos<<|;
    a[l]=a[r]=a[mid]=a[mid+]=lazy[pos]-;
    pre[pos]=suf[pos]=now[pos]=(a[l]==)*(r-l+);
    lazy[lpos]=lazy[rpos]=lazy[pos];
    lazy[pos]=;
}
void push_up(int l,int r,int pos)
{
    int mid=(l+r)>>;
    int lpos=pos<<,rpos=pos<<|;
    if (lazy[lpos])push_down(lson);
    if (lazy[rpos])push_down(rson);
    if (pre[lpos]+l==mid+)pre[pos]=pre[lpos]+pre[rpos];
    else pre[pos]=pre[lpos];
    if (r-suf[rpos]==mid)suf[pos]=suf[lpos]+suf[rpos];
    else suf[pos]=suf[rpos];
    now[pos]=max(now[lpos],now[rpos]);
    if (a[mid]==&&a[mid+]==)now[pos]=max(now[pos],suf[lpos]+pre[rpos]);
}
int getit(int l,int r,int pos)
{
    if (lazy[pos])push_down(l,r,pos);
    return now[pos];
}
void build(int l,int r,int pos)
{
    lazy[pos]=;
    if (l==r){
        a[l]=;
        suf[pos]=pre[pos]=now[pos]=;
        return ;
    }
    int mid=(l+r)>>;
    build(lson);
    build(rson);
    push_up(l,r,pos);
}
void upd(int l,int r,int pos,int l1,int r1,int v)
{
    if (lazy[pos])push_down(l,r,pos);
    if (l1<=l&&r1>=r){
        lazy[pos]=v;
        a[l]=a[r]=lazy[pos]-;
        pre[pos]=suf[pos]=now[pos]=(a[l]==)*(r-l+);
        return ;
    }
    int mid=(l+r)>>;
    if (mid>=l1)upd(lson,l1,r1,v);
    if (mid+<=r1)upd(rson,l1,r1,v);
    push_up(l,r,pos);
}
int query(int l,int r,int pos,int len)
{
    if (lazy[pos])push_down(l,r,pos);
    if (now[pos]<len)return ;
    if (l==r)return l;
    int mid=(l+r)>>;
    int lpos=pos<<,rpos=pos<<|;
    if (getit(lson)>=len)return query(lson,len);
    int v2=getit(rson);
    int v3=suf[lpos];
    if (a[mid]==&&a[mid+]==)v3=suf[lpos]+pre[rpos];
    if (v3>=len)return mid-suf[lpos]+;
    return query(rson,len);
}
int main()
{
    int n,m;
    int tr,x,y;
    while (~scanf("%d%d",&n,&m)){
        memset(a,,sizeof(a));
        memset(lazy,,sizeof(lazy));
        build(,n,);
        for (int i=;i<m;i++){
            scanf("%d%d",&tr,&x);
            if (tr==){
                int ans=query(,n,,x);
                printf("%d\n",ans);
                if (ans)
                upd(,n,,ans,ans+x-,);
            }else {
                scanf("%d",&y);
                upd(,n,,x,x+y-,);
            }
        }
    }
    return ;
}

* 在 vj 上看到了一个极快的vector暴力： 代码

* 分块这题应该也是可做的。（留坑，有空补上代码）

* 应该有瞎搞做法？（想到了补上）