Solution

线段树分治, 然后直接在线段树上dfs, 在进入/回溯的过程中维护并查集的merge/split.

对于split操作, 可以在merge时按秩合并, 然后利用栈记录, split时恢复即可.

Code

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
using namespace std;
#define rep(i,l,r) for(register int i=(l);i<=(r);++i)
#define repdo(i,l,r) for(register int i=(l);i>=(r);--i)
#define il inline
typedef double db;
typedef long long ll;//---------------------------------------
const int nsz=5050,msz=5e5+50;
int n,m;
int edno[nsz][nsz],pe=0,pq=0,pq1=1;
struct te{int f,t,l,r;}edge[msz];
struct tq{int a,b,t,ans;}que[msz];int fa[nsz],dep[nsz];
int stk[msz][2],top=0;//0 y; 1 dep[x]
int find(int p){return p==fa[p]?p:find(fa[p]);}
bool conn(int a,int b){return find(a)==find(b);}
void merge(int a,int b){a=find(a),b=find(b);if(a==b)return;if(dep[a]<dep[b])swap(a,b);stk[++top][0]=b,stk[top][1]=dep[a];fa[b]=a,dep[a]=max(dep[a],dep[b]+1);
}
void del(int top1){for(;top!=top1;--top){int a=stk[top][0],b=stk[top][1];dep[fa[a]]=b,fa[a]=a;}
}vector<int> ee[msz*4];
#define ls(p) ((p)<<1)
#define rs(p) ((p)<<1|1)
void insert(int v,int l,int r,int rt,int rl,int rr){if(l<=rl&&rr<=r){ee[rt].push_back(v);return;}int mid=(rl+rr)>>1;if(l<=mid)insert(v,l,r,ls(rt),rl,mid);if(mid<r)insert(v,l,r,rs(rt),mid+1,rr);
}
void dfs(int rt,int rl,int rr){int now=top;for(int i:ee[rt])merge(edge[i].f,edge[i].t);if(rl==rr){while(pq1<=pq&&que[pq1].t==rl)que[pq1].ans=conn(que[pq1].a,que[pq1].b),++pq1;}else{int mid=(rl+rr)>>1;dfs(ls(rt),rl,mid);dfs(rs(rt),mid+1,rr);}del(now);
}int main(){
//  freopen("a.in","r",stdin);
//  freopen("a.out","w",stdout);ios::sync_with_stdio(0),cin.tie(0);cin>>n>>m;int a,b,c;rep(i,1,m){cin>>a>>b>>c;if(b>c)swap(b,c);if(a==0){edge[++pe]=(te){b,c,i,m};edno[b][c]=pe;}else if(a==1){edge[edno[b][c]].r=i;}else{que[++pq]=(tq){b,c,i,0};}}rep(i,1,n)fa[i]=i,dep[i]=1;rep(i,1,pe)insert(i,edge[i].l,edge[i].r,1,1,m);dfs(1,1,m);rep(i,1,pq){cout<<(que[i].ans?"Y":"N")<<'\n';}return 0;
}