POJ3667 Hotel

最新推荐文章于 2019-07-24 20:47:51 发布

原创最新推荐文章于 2019-07-24 20:47:51 发布 · 503 阅读

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本文介绍了一种高效解决酒店入住与退房请求的算法，通过使用线段树数据结构来动态更新可用房间，确保每批客人能够获得连续且满意的住宿空间。详细阐述了算法的设计思路、实现过程及复杂度分析，旨在提高酒店运营效率并提升客户满意度。

Hotel

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 11154		Accepted: 4822

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

Sample Output

Source

USACO 2008 February Gold

题意：大概就是有几个人要住旅馆，要求房间必须是连续的，如果有的话，输出最小的房间编号，没有输出0，一共n个房间，编号1-n。
分析：线段树，区间更新，维护三个变量lsum,msum,rsum，lsum是包含最左边那个点的连续最多的空房，rsum是包含右边那个店的连续最多的空房，msum就是这个区间内最多的连续的空房。然后就是区间合并了。。。。。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX=50010;
struct seg
{
    int l,r;
    int lsum,rsum,msum;
    int mark;
    int m()
    {
        return (l+r)>>1;
    }
}tree[MAX*3];
void pushup(int k)
{
    int l_l=tree[2*k].r-tree[2*k].l+1;
    int r_l=tree[2*k+1].r-tree[2*k+1].l+1;
    tree[k].lsum=tree[2*k].lsum;
    if(tree[2*k].lsum==l_l)
        tree[k].lsum+=tree[2*k+1].lsum;
    tree[k].rsum=tree[2*k+1].rsum;
    if(tree[2*k+1].rsum==r_l)
        tree[k].rsum+=tree[2*k].rsum;
    tree[k].msum=max(max(tree[2*k].msum,tree[2*k+1].msum),tree[2*k].rsum+tree[2*k+1].lsum);
}
void pushdown(int k)
{
    int ll=tree[2*k].r-tree[2*k].l+1;
    int rr=tree[2*k+1].r-tree[2*k+1].l+1;
    tree[2*k].mark=tree[2*k+1].mark=tree[k].mark;
    tree[k].mark=-1;
    if(tree[2*k].mark)
        tree[2*k].lsum=tree[2*k].rsum=tree[2*k].msum=0;
    else
        tree[2*k].lsum=tree[2*k].rsum=tree[2*k].msum=ll;
    if(tree[2*k+1].mark)
        tree[2*k+1].lsum=tree[2*k+1].rsum=tree[2*k+1].msum=0;
    else
        tree[2*k+1].lsum=tree[2*k+1].rsum=tree[2*k+1].msum=rr;
}
void build(int l,int r,int k)
{
    tree[k].l=l;
    tree[k].r=r;
    tree[k].mark=-1;
    tree[k].lsum=tree[k].msum=tree[k].rsum=r-l+1;
    if(l==r)
        return;
    int mid=(l+r)>>1;
    build(l,mid,2*k);
    build(mid+1,r,2*k+1);
    pushup(k);
}
void update(int l,int r,int c,int k)
{
    if(tree[k].l==l&&tree[k].r==r)
    {
        tree[k].mark=c;
        if(c!=0)
            tree[k].lsum=tree[k].rsum=tree[k].msum=0;
        else
            tree[k].lsum=tree[k].rsum=tree[k].msum=r-l+1;
        return;
    }
    if(tree[k].mark!=-1)
        pushdown(k);
    int mid=tree[k].m();
    if(r<=mid)
        update(l,r,c,2*k);
    else if(l>mid)
        update(l,r,c,2*k+1);
    else
    {
        update(l,mid,c,2*k);
        update(mid+1,r,c,2*k+1);
    }
    pushup(k);
}
int query(int w,int k)
{
    if(tree[k].l==tree[k].r)
        return tree[k].l;
    if(tree[k].mark!=-1)
        pushdown(k);
    int mid=tree[k].m();
    if(tree[2*k].msum>=w)
        return query(w,2*k);
    else if(tree[2*k].rsum+tree[2*k+1].lsum>=w)
        return mid-tree[2*k].rsum+1;
    return query(w,2*k+1);
}
int main()
{
    int n,m,x,y,s;
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)==2)
    {
        build(1,n,1);
        while(m--)
        {
            scanf("%d",&s);
            if(s==1)
            {
                scanf("%d",&x);
                if(tree[1].msum<x)
                    printf("0\n");
                else
                {
                    int a=query(x,1);
                    printf("%d\n",a);
                    update(a,a+x-1,1,1);
                }
            }
            else
            {
                scanf("%d%d",&x,&y);
                update(x,x+y-1,0,1);
            }
        }
    }
    return 0;
}