```cpp
#include<bits/stdc++.h>
using namespace std;
#define vi vector<node >
#define pi pair<int ,int >
#define mcp(x,y) make_pair(x,y)
#define vii vector<int >
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
const int N = 4e5+9;
int n,m,q;
struct node{int u,v,w;};
map<pi ,int >mp,mp2;
vi e[N<<2];
struct que{int u,v,ans;}qp[N];
struct mod{int l,r;node x;}md[N];
set<pi >s;
void modify(int c,int l,int r,int x,int y,node el){
if(x<=l&&r<=y){e[c].push_back(el);return ;}
int mid=(l+r)>>1;
if(x<=mid) modify(ls(c),l,mid,x,y,el);if(y>mid) modify(rs(c),mid+1,r,x,y,el);
return ;
}
int f[N],xr[N],siz[N];
int find(int x){return f[x]==x?x:find(f[x]);}
int sum(int x){int sum=0;while(f[x]!=x) sum^=xr[x],x=f[x];sum^=xr[x];return sum;}
void add(vii &a,int x){
for(int i=30;i>=0;i--)
if((x>>i)&1){
if(!a[i]){a[i]=x;return ;}
else x^=a[i];
}
return ;
}
int query(vii a,int x){
int res=x;
for(int i=30;i>=0;i--) res=min(res,res^a[i]);
return res;
}
void dfs(int c,int l,int r,vii a){
vi st;
for(int i=0;i<e[c].size();i++){
int u=e[c][i].u,v=e[c][i].v,w=e[c][i].w;
int fu=find(u),fv=find(v);
if(fu!=fv){
if(siz[fu]>siz[fv]) swap(fu,fv);
st.push_back((node){fu,fv,xr[fu]});
f[fu]=fv,siz[fv]+=siz[fu],xr[fu]=xr[u]^w^xr[v];
}
else{int su=sum(u),sv=sum(v);add(a,su^sv^w);}
}
if(l==r){
int u=qp[l].u,v=qp[l].v;
int su=sum(u),sv=sum(v);qp[l].ans=query(a,su^sv);
return ;
}
int mid=(l+r)>>1;
dfs(ls(c),l,mid,a);dfs(rs(c),mid+1,r,a);
for(int i=st.size()-1;i>=0;i--){
int u=st[i].u,v=st[i].v,w=st[i].w;
f[u]=u,siz[v]-=siz[u],xr[u]=0;
}
return ;
}
int main(){
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++){
int u,v,w;scanf("%d%d%d",&u,&v,&w);
mp[mcp(u,v)]=1,mp[mcp(v,u)]=1;
mp2[mcp(u,v)]=w,mp2[mcp(v,u)]=w;
if(u>v) swap(u,v);
s.insert(mcp(u,v));
}
scanf("%d",&q);int tim=1,tot=0;
while(q--){
int op,u,v,w;scanf("%d%d%d",&op,&u,&v);
if(op==1){
scanf("%d",&w);
mp[mcp(u,v)]=tim,mp[mcp(v,u)]=tim;
mp2[mcp(u,v)]=w,mp2[mcp(v,u)]=w;
if(u>v) swap(u,v);
s.insert(mcp(u,v));
}
if(op==2){
int l=mp[mcp(u,v)],r=tim-1,w=mp2[mcp(u,v)];
if(u>v) swap(u,v);s.erase(mcp(u,v));
if(l>r) continue;
md[++tot]=(mod){l,r,(node){u,v,w}};
}
if(op==3){
qp[tim].u=u,qp[tim].v=v;tim++;
}
}
tim--;
for(set<pi >::iterator it=s.begin();it!=s.end();++it){
int l=mp[(*it)],r=tim;node x=(node){(*it).first,(*it).second,mp2[(*it)]};
if(l<=r) md[++tot]=(mod){l,r,x};
}
// for(int i=1;i<=tot;i++) printf("[%d,%d] %d -> %d length %d\n",md[i].l,md[i].r,md[i].x.u,md[i].x.v,md[i].x.w);
for(int i=1;i<=tot;i++)
modify(1,1,tim,md[i].l,md[i].r,md[i].x);
for(int i=1;i<=n;i++) f[i]=i,xr[i]=0,siz[i]=1;
vii a;a.resize(32);
dfs(1,1,tim,a);
for(int i=1;i<=tim;i++) printf("%d\n",qp[i].ans);
return 0;
}
```
by wuhao2005 @ 2021-02-19 16:09:49
还有顺便问下这题数据怎么造比较快
(感觉加边删边还要维护连通性比较麻烦)
by wuhao2005 @ 2021-02-19 16:16:38
膜拜!
by Oakenshield @ 2021-02-19 16:17:14
膜拜!
by Arkadyevna @ 2021-02-19 16:28:52
膜拜!
by DPair @ 2021-02-19 16:35:50
不得不说,线段树分治的常数是真心大啊
告诫后人:本题中按秩合并的并查集每个点记录的是到其并查集上父亲的异或路径长度,而两个并查集合并时,两边的根节点更新时用的是 $u,v$ 到根节点的异或路径长度与 $w$ 的异或和,需要两边的点遍历的根节点得到,而不是直接取该点并查集中记录的(~~尽管我认为没有人和我犯同样的错误~~)
```cpp
if(fu!=fv){
if(siz[fu]>siz[fv]) swap(fu,fv);
st.push_back((node){fu,fv,xr[fu]});
f[fu]=fv,siz[fv]+=siz[fu],xr[fu]=sum(u)^w^sum(v);
}
else{int su=sum(u),sv=sum(v);add(a,su^sv^w);
//而不是原先代码里的add(a,xr[u]^xr[v]^w);
}
```
当然,好像用 LCT 写就没这个问题
by wuhao2005 @ 2021-02-19 18:40:25
@[wuhao2005](/user/81274) 好了现在有一个了
by fanypcd @ 2022-01-24 18:14:42