Description
Input
Output
Sample Input
! 1
! 2
+ 1 2
! 1
! 2
- 1 2
! 1
! 2
Sample Output
只有第4和第5條記錄對應的消息被看到過。其他消息發送時,1和2不是好友。
對100%的數據,N<=200000,M<=500000
據說正解是set維護每個人的朋友,然後考慮每次加邊、刪邊對答案的貢獻......
作爲腦子有坑的選手,我打了一棵帶標記Treap...
每個人維護一棵Treap,+ x y則在x的Treap裏插入y,-則刪去,! x則給x的Treap打上+1標記,最後暴力將所有標記下放,更新答案
代碼
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#define maxn 200005
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
inline int read()
{ char c=getchar();int x=0,y=1;
while(c<'0'||c>'9'){if(c=='-') y=-1;c=getchar();}
while(c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*y;
}
inline char readchar(){char c;for(c=getchar();c!='!'&&c!='-'&&c!='+';c=getchar());return c;}
int n,m,ans[maxn];
struct Treap
{ Treap *ch[2];int s,id,ad,r;
Treap(int x=0):id(x),ad(0),s(1),r(rand()){ch[0]=ch[1]=NULL;}
inline int sz(){return this?this->s:0;}
inline void addv(int x){if(!this) return;this->ad+=x;ans[this->id]+=x;}
inline void mt(){this->s=ch[0]->sz()+ch[1]->sz()+1;}
inline void dn(){if(this->ad){ch[0]->addv(ad);ch[1]->addv(ad);this->ad=0;}}
}*root[maxn];
inline void rotate(Treap*& x,int d)
{ Treap* k=x->ch[d^1];x->dn();k->dn();
x->ch[d^1]=k->ch[d];k->ch[d]=x;
x->mt();k->mt();x=k;
}
inline void insert(Treap*& x,int id)
{ if(!x){x=new Treap(id);return;}
x->dn();int d=x->id<id;
insert(x->ch[d],id);
if(x->ch[d]->r<x->r) rotate(x,d^1);
x->mt();
}
inline void del(Treap*& x,int id)
{ x->dn();
if(x->id==id)
{ if(x->ch[0]&&x->ch[1])
{ int d=x->ch[0]->r<x->ch[1]->r;
rotate(x,d);del(x->ch[d],id);
}
else
{ Treap* tmp=x;x=x->ch[0]?x->ch[0]:x->ch[1];
delete tmp;
}
}
else del(x->ch[x->id<id],id);
}
inline void remove(Treap* x)
{ if(!x) return;
x->dn();remove(x->ch[0]);remove(x->ch[1]);
delete x;
}
int main()
{ n=read();m=read();int x,y;char ord;
for(int i=1;i<=m;++i)
{ ord=readchar();
if(ord=='!'){x=read();root[x]->addv(1);}
else if(ord=='+'){x=read();y=read();insert(root[x],y);insert(root[y],x);}
else{x=read();y=read();del(root[x],y);del(root[y],x);}
}
for(int i=1;i<=n;++i) remove(root[i]);
printf("%d",ans[1]);
for(int i=2;i<=n;++i) printf(" %d",ans[i]);
return 0;
}