[USACO13JAN] Seating

https://www.luogu.org/problem/show?pid=3071

题目描述

To earn some extra money, the cows have opened a restaurant in their barn specializing in milkshakes. The restaurant has N seats (1 <= N <= 500,000) in a row. Initially, they are all empty.

Throughout the day, there are M different events that happen in sequence at the restaurant (1 <= M <= 300,000). The two types of events that can happen are:

1. A party of size p arrives (1 <= p <= N). Bessie wants to seat the party in a contiguous block of p empty seats. If this is possible, she does so in the lowest position possible in the list of seats. If it is impossible, the party is turned away.

2. A range [a,b] is given (1 <= a <= b <= N), and everybody in that range of seats leaves.

Please help Bessie count the total number of parties that are turned away over the course of the day.

有一排n个座位，m次操作。A操作：将a名客人安置到最左的连续a个空位中，没有则不操作。L操作：[a,b]的客人离开。

求A操作的失败次数。

输入输出格式

输入格式：

• Line 1: Two space-separated integers, N and M.

• Lines 2..M+1: Each line describes a single event. It is either a line of the form "A p" (meaning a party of size p arrives) or "L a b" (meaning that all cows in the range [a, b] leave).

输出格式：

• Line 1: The number of parties that are turned away.

输入输出样例

输入样例#1：

10 4
A 6
L 2 4
A 5
A 2

输出样例#1：

1 

线段树
实践再次证明：数组比结构体要快

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,x,y,opl,opr;
#define N 500001*4
int maxx[N],max_l[N],max_r[N],sum[N],flag[N];
void build(int k,int l,int r)
{
    sum[k]=maxx[k]=max_l[k]=max_r[k]=r-l+;
    if(l==r) return;
    int mid=l+r>>;
    build(k<<,l,mid);build((k<<)+,mid+,r);
}
void down(int k)
{
    flag[k<<]=flag[(k<<)+]=flag[k];
    if(flag[k]==)
        maxx[k<<]=max_l[k<<]=max_r[k<<]=maxx[(k<<)+]=max_l[(k<<)+]=max_r[(k<<)+]=;
    else
    {
        maxx[k<<]=max_l[k<<]=max_r[k<<]=sum[k<<];
        maxx[(k<<)+]=max_l[(k<<)+]=max_r[(k<<)+]=sum[(k<<)+];
    }
    flag[k]=;
}
int ask(int k,int l,int r)
{
    if(l==r) return l;
    if(flag[k]) down(k);
    int mid=l+r>>;
    if(maxx[k<<]>=x) return ask(k<<,l,mid);
    if(max_l[(k<<)+]+max_r[k<<]>=x) return mid-max_r[k<<]+;
    return ask((k<<)+,mid+,r);
}
void up(int k)
{
    maxx[k]=max(max(maxx[k<<],maxx[(k<<)+]),max_r[k<<]+max_l[(k<<)+]);
    if(sum[k<<]==maxx[k<<])  max_l[k]=sum[k<<]+max_l[(k<<)+];
    else max_l[k]=max_l[k<<];
    if(sum[(k<<)+]==maxx[(k<<)+]) max_r[k]=sum[(k<<)+]+max_r[k<<];
    else max_r[k]=max_r[(k<<)+];
}
void change(int k,int f,int l,int r)
{
    if(l>=opl&&r<=opr)
    {
        if(f==)
        {
            maxx[k]=max_l[k]=max_r[k]=;
            flag[k]=;
            return;
        }
        else
        {
            maxx[k]=max_l[k]=max_r[k]=sum[k];
            flag[k]=;
            return;
        }
    }
    if(flag[k]) down(k);
    int mid=l+r>>;
    if(opr<=mid) change(k<<,f,l,mid);
    else if(opl>mid) change((k<<)+,f,mid+,r);
    else
    {
        change(k<<,f,l,mid);
        change((k<<)+,f,mid+,r);
    }
    up(k);
}
void read(int &x)
{
    x=; char c=getchar();
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) { x=x*+c-''; c=getchar(); }
}
int main()
{
    int cnt=;
    char p[];
    read(n); read(m);
    build(,,n);
    for(int i=;i<=m;i++)
    {
        scanf("%s",p);
        if(p[]=='A')
        {
            read(x);
            if(maxx[]<x) cnt++;
            else
            {
                opl=ask(,,n);
                opr=opl+x-;
                change(,,,n);
            }
        }
        else
        {
            read(opl); read(opr);
            change(,,,n);
        }
    }
    printf("%d",cnt);
}