problem

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

思路

簡化題意即為：

兩種操作
1.找是否存在指定長度的空的（未標記的）區間，如果有，輸出最左端的點，並將這個區間標記
2.從一指定點開始，消除一段指定長度的區間的標記

看到這題很容易想到線段樹，其常用來維護區間的資訊。
其中操作2消除標記就是最基本的區間覆蓋
關鍵問題是操作1怎麼處理
如果只是讓判斷是否存這樣連續未標記的區間，那我只要再維護一個數組，比如slen,它表示當前結點最大連續未標記長度。在pushup和pushdown時維護一下即可。
令人頭疼的是，它不僅要判斷是否存在足夠長度未標記的區間，還要找出最左端的位置。

這時候我沒想到怎麼做，看了一下別人的思想，頓有醍醐灌頂之感

維護三個資訊，分別是:
lsum 當前區間從左端點開始向又看，最大連續未標記區間是多長
rsum 當前區間從右端點開始向左看，最大連續未標記區間是多長
msum 遞迴定義：max{ 左半區間msum , 右半區間msum , 左半區間的rsum+右半區間lsum(這兩部分正好拼在一起)}
比如區間6~10
標記是10011 那lsum=0 rsum=0 msum=2 ;
標記是00111 那lsum=2 rsum=0 msum=2 ;

這樣定義的msum即為當前結點表示區間內的最大未標記連續區間

對於lsum和rsum的維護，核心程式碼為：

lsum[rt]=lsum[rt<<1];//當前區間左大連續自然等於左兒子左大連續
rsum[rt]=rsum[rt<<1|1];//當前區間右大連續自然等於右兒子右大連續
//但這樣是極限嗎？顯然不是 如果左兒子全未標記 可能右兒子的左部分也未標記 就有了下面的程式碼
if(lsum[rt]==len-(len>>1)) lsum[rt]+=lsum[rt<<1|1];//if條件是：左大連續等於左區間長 加上右兒子左大連續
if(rsum[rt]==(len>>1)) rsum[rt]+=rsum[rt<<1];//同理

程式碼示例

#include<iostream>
#include<cstdio>
#include<algorithm>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define bug cout<<"test"<<endl
using namespace std;
const int maxn=50100;

int lsum[maxn<<2],rsum[maxn<<2],msum[maxn<<2];
int cov[maxn<<2];//-1 0 1三種狀態

void PushDown(int rt,int len){
    if(cov[rt]!=-1){
        cov[rt<<1]=cov[rt<<1|1]=cov[rt];
        msum[rt<<1]=lsum[rt<<1]=rsum[rt<<1]=cov[rt]?0:len-(len>>1);
        msum[rt<<1|1]=lsum[rt<<1|1]=rsum[rt<<1|1]=cov[rt]?0:len>>1;
        cov[rt]=-1;
    }
}

void PushUp(int rt,int len){
    lsum[rt]=lsum[rt<<1];
    rsum[rt]=rsum[rt<<1|1];
    if(lsum[rt]==len-(len>>1)) lsum[rt]+=lsum[rt<<1|1];
    if(rsum[rt]==(len>>1)) rsum[rt]+=rsum[rt<<1];
    msum[rt]=max(max(msum[rt<<1],msum[rt<<1|1]),rsum[rt<<1]+lsum[rt<<1|1]);
}

void build(int l,int r,int rt){
    msum[rt]=lsum[rt]=rsum[rt]=r-l+1;//一開始均為區間長度
    cov[rt]=-1;
    if(l==r) return ;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}

void update(int L,int R,int c,int l,int r,int rt){
    if(L<=l&&r<=R){
        msum[rt]=lsum[rt]=rsum[rt]=c?0:r-l+1;
        cov[rt]=c;
        return ;
    }
    PushDown(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m) update(L,R,c,lson);
    if(m<R) update(L,R,c,rson);
    PushUp(rt,r-l+1);
}

int query(int w,int l,int r,int rt){
    if(l==r) return l;
    PushDown(rt,r-l+1);
    int m=(l+r)>>1;
    if(msum[rt<<1]>=w) return query(w,lson);//先考慮左兒子
    else if(rsum[rt<<1]+lsum[rt<<1|1]>=w) return m-rsum[rt<<1]+1;//再考慮兩部分合並的中間部分
    return query(w,rson);//最後考慮右兒子
}

int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    build(1,n,1);
    while(m--){
        int op,a,b;
        scanf("%d",&op);
        if(op==1){
            scanf("%d",&a);
            if(msum[1]<a) puts("0");//最大空出長度都不滿足
            else{
                int p=query(a,1,n,1);
                printf("%d\n",p);
                update(p,p+a-1,1,1,n,1);
            }
        }

        else{
            scanf("%d %d",&a,&b);
            update(a,a+b-1,0,1,n,1);
        }
    }
    return 0;
}