Week 8.8---8.14

It's a new week, eh.

8.8

Do CF

After a day's game, B thought well and complicated. Always count SG is wrong.

C is not understand how to deal with the first T-seconds inside not be 2 operation of the number of operations dropped

CF 366 Div2 c C-thor

To each node A queue, the queue is stored in the corresponding to the time it added,

In the 3 operation, you can maintain a position, if the last operation to T1 then this operation will only need to T1 to T2.

I didn't expect to throw every time as a thing into the queue.

1#include <stdio.h>2#include <string.h>3#include <stdlib.h>4#include <time.h>5#include <math.h>6#include <vector>7#include <map>8#include <Set>9#include <stack>Ten#include <queue> One#include <string> A#include <iostream> -#include <algorithm> - using namespacestd; the  -typedefLong LongLL; -typedef pair<int,int>PII; -typedef pair<Double,Double>PDD; + Const DoubleEPS = 1e-8; - Const intINF = (1<< -) -1; + Const intMAXN =300010; A intN,M,QUE[MAXN]; atqueue<int>Q[MAXN]; -  - voidsolve () { -     intOp,x,t =0, ans =0, pos =0; -memset (que,-1,sizeof(que)); -      for(inti =1; I <= m;i++){ inscanf"%d%d",&op,&x); -         if(OP = =1){ toQUE[++T] =x; + Q[x].push (t); -ans++; the         } *         Else if(OP = =2){ $              while(!Q[x].empty ()) {Panax Notoginseng                 intU =Q[x].front (); Q[x].pop (); -Que[u] =-1; theans--; +             } A         } the         Else{ +             inttop =min (t,x); -              for(inti = pos+1; I <= top;i++){ $                 if(Que[i] = =-1)Continue; $                 intv =Que[i]; - Q[v].pop (); -ans--; the             } -pos =Max (pos,top);Wuyi         } theprintf"%d\n", ans); -     } Wu } -  About intMain () { $scanf"%d%d",&n,&m); - solve (); -     return 0; -}

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Week 8.8---8.14