Codeforces Round #366 (Div. 1) 题解 (待补)

Thor

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int n,q;
#define
int l[MAXN],r[MAXN];
vi a[MAXN];
int b[MAXN]={0};
int main()
{
//  freopen("A.in","r",stdin);
//  freopen(".out","w",stdout);
    n=read(),q=read();
    MEM(a) MEM(l) MEM(r)
    int siz=1,ans=0; l[1]=0; r[0]=1;
    int idx=1; 
    For(i,q) {
        int p=read(),c=read();
        if (p==1) {
            siz++; b[siz-1]=1;
            a[c].pb(siz-1); ++ans;
        } else if (p==2) {
            int sz=a[c].size();
            Rep(j,sz) {
                int p=a[c][j];
                if (!b[p]) continue;
                b[p]=0;
                --ans;
            }
            a[c].clear();
        } else {
            Fork(j,idx,c) {
                if (b[j]) {
                    b[j]=0; ans--;
                }
            }
            idx=max(idx,c);
        }
        cout<<ans<<endl;
    }

    return 0;
}

Ant Man