Codeforces Round #366 (Div. 2) C list

链接：戳这里

C. 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).

题意：

手机app应用通知，三种操作

1 x：x应用通知了一条信息

2 x：阅读了所有x应用的通知信息

3 x：操作1加入的前x条信息已经阅读(通过操作2阅读过的这里要继续阅读

思路：

链表模拟,记录两个权值，当前是操作1，应用程序x插入信息的时间以及属于应用x的信息

对于操作2：减去应用程序x未读的数量，存下当前应用程序已经看到了第cnt条信息

操作3：当前阅读的x如果<=链表头节点的插入时间，直接break。如果>插入时间，判断是否x程序是否在操作2后，也就是pre[x]之后，是的话减去阅读量1

链表删除3操作下阅读的前t个通知

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<list>#include<iomanip>#include<cmath>#include<bitset>#define mst(ss,b) memset((ss),(b),sizeof(ss))///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;#define INF (1ll<<60)-1#define Max 1e9using namespace std;int n,q,t,x;list<pair<int,int> > anw;int pre[300100],num[300100];int main(){    int cnt=0;    scanf("%d%d",&n,&q);    int ans=0,L=1;    for(int i=1;i<=q;i++){        scanf("%d%d",&t,&x);        if(t==1){            anw.push_back(make_pair(++cnt,x));            ans++;            num[x]++;        } else if(t==2){            ans-=num[x];            pre[x]=cnt;            num[x]=0;        } else {            list<pair<int,int> >::iterator it;            for(it=anw.begin();it!=anw.end();){                if(it->first > x) break;                if(it->first > pre[it->second]){                    ans--;                    num[it->second]--;                }                anw.erase(it++);            }        }        printf("%d\n",ans);    }    return 0;}