| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 9597 | Accepted: 4113 |
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 Di (1 ≤
Di ≤ N) and approach the front desk to check in. Each group
i requests a set of Di contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers
r..r+Di-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 Xi and
Di which specify the vacating of rooms Xi ..Xi +Di-1 (1 ≤
Xi ≤ N-Di+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
Di (b) Three space-separated integers representing a check-out: 2,
Xi, and Di
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
num数组记录该区间是否住慢，0代表都是空的，1代表都住了人。-1代表有的住了，有的空着。
d[rt][0]代表从区间最左端开始的空的最长长度，d[rt][1]代表从区间最右端开始的空的最长长度，d[rt][2]代表区间的最长空的长度。
#include<cstdio>
#define maxn 51111
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Max(a,b) (a>b?a:b)
int num[maxn<<2],d[maxn<<2][3];
void PushDown(int rt,int tmp)
{
if(num[rt]==-1)
return ;
num[rt<<1]=num[rt<<1|1]=num[rt];
if(num[rt]==0)
{
d[rt<<1][0]=d[rt<<1][1]=d[rt<<1][2]=(tmp+1)>>1;
d[rt<<1|1][0]=d[rt<<1|1][1]=d[rt<<1|1][2]=tmp>>1;
}
else
{
d[rt<<1][0]=d[rt<<1][1]=d[rt<<1][2]=0;
d[rt<<1|1][0]=d[rt<<1|1][1]=d[rt<<1|1][2]=0;
}
num[rt]=-1;
}
void build(int l,int r,int rt)
{
num[rt]=0;
for(int i=0;i<3;i++)
d[rt][i]=r-l+1;
if(l==r)
return ;
int m=(l+r)>>1;
build(lson);
build(rson);
}
int check(int l,int r,int rt,int D) //查找满足条件的最左点
{
if(d[rt][2]<D)
return 0;
PushDown(rt,r-l+1);
int m=(l+r)>>1;
if(d[rt<<1][2]>=D)
return check(lson,D);
if(d[rt<<1][1]+d[rt<<1|1][0]>=D)
return m-d[rt<<1][1]+1;
return check(rson,D);
}
void PushUp(int rt,int tmp)
{
if(d[rt<<1][0] < (tmp+1)>>1)
d[rt][0]=d[rt<<1][0];
else
d[rt][0]=((tmp+1)>>1)+d[rt<<1|1][0];
if(d[rt<<1|1][1]<tmp>>1)
d[rt][1]=d[rt<<1|1][1];
else
d[rt][1]=d[rt<<1][1]+(tmp>>1);
d[rt][2]=Max(d[rt<<1][2],d[rt<<1|1][2]);
d[rt][2]=Max(d[rt][2],d[rt<<1][1]+d[rt<<1|1][0]);
}
void updata(int l,int r,int rt,int L,int R,int flag)
{
if(l>=L && r<=R)
{
num[rt]=flag;
if(flag==1)
d[rt][0]=d[rt][1]=d[rt][2]=0;
else
d[rt][0]=d[rt][1]=d[rt][2]=r-l+1;
return ;
}
PushDown(rt,r-l+1);
int m=(l+r)>>1;
if(L<=m)
updata(lson,L,R,flag);
if(R>m)
updata(rson,L,R,flag);
PushUp(rt,r-l+1);
}
int main()
{
int N,M;
int flag,X,D,tmp;
while(scanf("%d%d",&N,&M)==2)
{
build(1,N,1);
while(M--)
{
scanf("%d",&flag);
if(flag==1)
{
scanf("%d",&D);
tmp=check(1,N,1,D);
printf("%d\n",tmp);
if(tmp!=0)
{
updata(1,N,1,tmp,tmp+D-1,1);
}
}
if(flag==2)
{
scanf("%d%d",&X,&D);
updata(1,N,1,X,X+D-1,0);
}
}
}
return 0;
}