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A. Thor
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Thor is getting used to the Earth. As a gift Loki gave him a smartphone. There are n applications on this phone. Thor is fascinated by this phone. He has only one minor issue: he can’t count the number of unread notifications generated by those applications (maybe Loki put a curse on it so he can’t).

q events are about to happen (in chronological order). They are of three types:

Application x generates a notification (this new notification is unread).
Thor reads all notifications generated so far by application x (he may re-read some notifications).
Thor reads the first t notifications generated by phone applications (notifications generated in first t events of the first type). It’s guaranteed that there were at least t events of the first type before this event. Please note that he doesn’t read first t unread notifications, he just reads the very first t notifications generated on his phone and he may re-read some of them in this operation.
Please help Thor and tell him the number of unread notifications after each event. You may assume that initially there are no notifications in the phone.

Input
The first line of input contains two integers n and q (1 ≤ n, q ≤ 300 000) — the number of applications and the number of events to happen.

The next q lines contain the events. The i-th of these lines starts with an integer typei — type of the i-th event. If typei = 1 or typei = 2 then it is followed by an integer xi. Otherwise it is followed by an integer ti (1 ≤ typei ≤ 3, 1 ≤ xi ≤ n, 1 ≤ ti ≤ q).

Output
Print the number of unread notifications after each event.

Examples
input
3 4
1 3
1 1
1 2
2 3
output
1
2
3
2
input
4 6
1 2
1 4
1 2
3 3
1 3
1 3
output
1
2
3
0
1
2
Note
In the first sample:

Application 3 generates a notification (there is 1 unread notification).
Application 1 generates a notification (there are 2 unread notifications).
Application 2 generates a notification (there are 3 unread notifications).
Thor reads the notification generated by application 3, there are 2 unread notifications left.
In the second sample test:

Application 2 generates a notification (there is 1 unread notification).
Application 4 generates a notification (there are 2 unread notifications).
Application 2 generates a notification (there are 3 unread notifications).
Thor reads first three notifications and since there are only three of them so far, there will be no unread notification left.
Application 3 generates a notification (there is 1 unread notification).
Application 3 generates a notification (there are 2 unread notifications).

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題目大意 :
就是你有一個手機 有三種操作
1,x軟體收到一個資訊
2,看了所有x軟體的資訊
3,看了第1~第x次收到的資訊 （被看過的也算）

每次都輸出一下當前手機裡的未讀資訊的個數

題解 ：
首先把資訊編號
用一個vector 和set 維護下就好了
vector有n個 代表n個軟體 每次向n[x]中加入新的資訊編號
set儲存所有的資訊

維護的時候
對於1 操作 把資訊編號加入vector 和set
對於2 操作 對應n[x]遍歷一遍 從set中刪除
對於3 操作 遍歷1~x 從set中刪除 （這裡注意每次的x要記錄一下 然後下一次遍歷的時候只要遍歷這個x到下一個x的區間就行了 之前被刪除的不用再刪除一遍 否則會TLE）
每次輸出set的大小就行了

這樣總體複雜度是O(2n)

/****** 這些shi牢騷
這道題是賽後補得 並且看了網上的題解
當時 想到了思路 當時沒有做主要是怕TLE 在一個不太會用set （我是小白） 當時想到的是對陣列二分查詢整個區間 然後刪除 當對判斷資料的統計這一塊只想到了線段樹（賽前幾天才學習資料結構，做什麼題都想用線段樹） 然後旁邊隊友 風騷的A了這道題後 告訴用STL做 然而對於不會STL的我老說 根本沒法搞。 So我就不做了。。。
仔細想來 根本不用那麼麻煩 多熟練下vector map set 這東西其實是能秒的。。
******/
附本題程式碼
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#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 300010;

vector<int> app[maxn];
set<int> cnt;

int main()
{
    int n,q,x,type,num;
    while(~scanf("%d%d",&n,&q))
    {
        num = 0;
        cnt.clear();
        int last = 0;
        for(int i=0;i<q;i++)
        {
            scanf("%d%d",&type,&x);
            if(type==1)
            {
                app[x].push_back(++num);
                cnt.insert(num);
            }
            else if(type==2)
            {
                for(int i=0;i<app[x].size();i++)
                    cnt.erase(app[x][i]);
                app[x].clear();
            }
            else
            {
                for(int i=last;i<=x;i++)
                {
                    cnt.erase(i);
                }
                last = max(x,last);
            }
            printf("%d\n",cnt.size());
        }
    }
    return 0;
}