SDNU_ACM_ICPC_2020_Winter_Practice_1st F 队列 pair

题目

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).

大意：
手机的app会发通知消息，有三种操作：
①某个app发出一条消息
②阅读某个app的所有消息
③阅读前num条未读消息（注意是全部消息的前num条，有些消息会重读）

思路

num 用来记录一共产生了几条消息（不减去已读消息）
队列 queue f[30008]; 记录app产生的是当前的第几条消息，并且只保留未读消息的序号
队列 queue<pair<int,int>>road 记录第几条消息是有哪个app产生的
数组 bool vis[300008]; 用来判断第几条消息是否已读。
sum 记录有几条未读消息

代码

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
using namespace std;
queue <int> f[300008];
queue<pair<int,int> >road;
bool vis[300008];
int main()
{
    int a,b,last=0,sum=0,n,x,m;
    cin>>n>>m;
    int num=0;
    while(m--)
    {
        scanf("%d %d",&a,&b);
        if(a==1)
        {
            f[b].push(++num);
            road.push(make_pair(num,b));///b  app产生了第num条消息
            sum++;
        }
        else if(a==2)
        {
            while(!f[b].empty())
            {
                if(!vis[f[b].front()])///这条消息未读
                {
                    sum--;
                    vis[f[b].front()]=true;///标记已读
                }
                f[b].pop();///删除
            }
        }
        else
        {
            while(!road.empty()&&road.front().first<=b)
            {
                if(!vis[road.front().first])///这条消息未读
                {
                    sum--;
                    vis[road.front().first]=true;///标记已读
                    f[road.front().second].pop();///  这个app 中的这条消息删掉
                }
                road.pop();  ///删掉处理过的消息
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}