bzoj 3050: [Usaco2013 Jan]Seating

3050: [Usaco2013 Jan]Seating

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 128  Solved: 73
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Description

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.

m(m<=300,000)个操作。操作分2种：

1.A p，表示把编号最小的空着的长度为p的区间图上颜色。

2.L a b，表示把从a到b的区间（包括端点）全部擦干净（没颜色还是没颜色）。

Q：有多少个操作1不能实现？

 

Input

 

* 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).

Output

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

 

Sample Input

10 4
A 6
L 2 4
A 5
A 2
INPUT DETAILS: There are 10 seats, and 4 events. First, a party of 6 cows arrives. Then all cows in seats 2..4 depart. Next, a party of 5 arrives, followed by a party of 2.

Sample Output

1

OUTPUT DETAILS: Party #3 is turned away. All other parties are seated.

HINT

样例解释：

  首先将1~6图上颜色，再把2~4擦掉，这时不存在长度为5的空着的区间，该操作不能实现，跳过。最后把2~3图上颜色。所以不能实现的操作1有一个，即第三个操作。

Source

Gold

题解参见：poj 3667 hotel 

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define N 500003using namespace std;int n,m,ans;int sum[4*N],lx[N*4],rx[N*4],delta[4*N];void update(int x){  sum[x]=sum[x<<1]+sum[x<<1|1];  lx[x]=rx[x]=sum[x];}void build(int now,int l,int r){  if (l==r)  {    sum[now]=lx[now]=rx[now]=1;    return;  }  int mid=(l+r)/2;  build(now<<1,l,mid);  build(now<<1|1,mid+1,r);  update(now);}void pushdown(int x,int l,int r){  if (!delta[x]) return;  int mid=(l+r)/2;  sum[x<<1]=lx[x<<1]=rx[x<<1]=(delta[x]==1?0:mid-l+1);  sum[x<<1|1]=lx[x<<1|1]=rx[x<<1|1]=(delta[x]==1?0:r-mid);  delta[x<<1]=delta[x<<1|1]=delta[x];  delta[x]=0;}int find(int now,int l,int r,int x){  if(l==r) return l;  pushdown(now,l,r);  int mid=(l+r)/2;  if (sum[now<<1]>=x)   return find(now<<1,l,mid,x);  else  if (rx[now<<1]+lx[now<<1|1]>=x)   return mid-rx[now<<1]+1;  else   return find(now<<1|1,mid+1,r,x);}void update1(int x,int l,int r){  int mid=(l+r)/2;  sum[x]=max(max(sum[x<<1],sum[x<<1|1]),rx[x<<1]+lx[x<<1|1]);  lx[x]=lx[x<<1]+(lx[x<<1]==mid-l+1?lx[x<<1|1]:0);  rx[x]=rx[x<<1|1]+(rx[x<<1|1]==r-mid?rx[x<<1]:0);}void qjchange(int now,int l,int r,int ll,int rr,int v){  if (l>=ll&&r<=rr)  {  if (v==0)   sum[now]=lx[now]=rx[now]=0;  else   sum[now]=lx[now]=rx[now]=r-l+1;  delta[now]=v+1;  return;  }  pushdown(now,l,r);  int mid=(l+r)/2;  if (ll<=mid)   qjchange(now<<1,l,mid,ll,rr,v);  if (rr>mid)   qjchange(now<<1|1,mid+1,r,ll,rr,v);  update1(now,l,r);}int main(){  freopen("seating.in","r",stdin);  freopen("seating.out","w",stdout);  scanf("%d%d",&n,&m);  build(1,1,n);  for (int i=1;i<=m;i++)  {   char s[10]; scanf("%s",s);   if (s[0]=='A')   {   int x; scanf("%d",&x);   if (sum[1]<x)    ans++;   else    {      int pos=find(1,1,n,x);      qjchange(1,1,n,pos,pos+x-1,0);    }   }   else    {      int x,y; scanf("%d%d",&x,&y);      qjchange(1,1,n,x,y,1);    }  }  printf("%d\n",ans);}