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正解：LCT維護動態最小生成樹。

這題就是要動態維護經過x到y的路徑的最大值的最小值。那么我們可以先考慮離線做法，即倒著操作，因為刪邊以后沒辦法維護最值了。一開始先把沒有限制的邊加上，然后倒著詢問時直接把邊加上就行了。不難看出這題是要維護最小生成樹，那么當加入一條邊時如果兩個點已經連通，就要把連通他們的最大的邊刪掉。所以這道題可以用LCT來動態維護添邊刪邊的操作。于是這題就能解決了（不過加強版卡空間woc）。

//It is made by wfj_2048~#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <vector>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>#define N (1100010)#define inf (1<<30)#define il inline#define RG register#define ll long long#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)using namespace std;map <int,int> hsh[N];struct edge{ int x,y,w; }g[N],q[N];int ch[N][2],fa[N],mx[N],sum[N],val[N],lazy[N],st[N],vis[N],ans[N],n,m,Q;il int gi(){    RG int x=0,q=1; RG char ch=getchar(); while ((ch<'0' || ch>'9') && ch!='-') ch=getchar();    if (ch=='-') q=-1,ch=getchar(); while (ch>='0' && ch<='9') x=x*10+ch-48,ch=getchar(); return q*x;}il int cmp(const edge &a,const edge &b){ return a.w<b.w; }il void pushdown(RG int x){ lazy[x]=0,lazy[ch[x][0]]^=1,lazy[ch[x][1]]^=1,swap(ch[x][0],ch[x][1]); return; }il void pushup(RG int x){    RG int t=mx[ch[x][mx[ch[x][1]]>mx[ch[x][0]]]];    mx[x]=(val[t]<val[x] ? x : t); return;}il int isroot(RG int x){ return ch[fa[x]][0]!=x && ch[fa[x]][1]!=x; }il void rotate(RG int x){    RG int y=fa[x],z=fa[y],k=ch[y][0]==x; if (!isroot(y)) ch[z][ch[z][1]==y]=x; fa[x]=z;    ch[y][k^1]=ch[x][k],fa[ch[x][k]]=y,ch[x][k]=y,fa[y]=x,pushup(y),pushup(x); return;}il void splay(RG int x){    RG int top=0; st[++top]=x;    for (RG int i=x;!isroot(i);i=fa[i]) st[++top]=fa[i];    for (RG int i=top;i;--i) if (lazy[st[i]]) pushdown(st[i]);    while (!isroot(x)){	RG int y=fa[x],z=fa[y];	if (!isroot(y)){	    if ((ch[z][0]==y)^(ch[y][0]==x)) rotate(x); else rotate(y);	}	rotate(x);    }    return;}il void access(RG int x){ RG int t=0; while (x) splay(x),ch[x][1]=t,pushup(x),t=x,x=fa[x]; return; }il void makeroot(RG int x){ access(x),splay(x),lazy[x]^=1; return; }il void link(RG int x,RG int y){ makeroot(x),fa[x]=y; return; }il void cut(RG int x,RG int y){ makeroot(x),access(y),splay(y),ch[y][0]=fa[x]=0; return; }il int querymax(RG int x,RG int y){ makeroot(x),access(y),splay(y); return mx[y]; }il int find(RG int x){ access(x),splay(x); while (ch[x][0]) x=ch[x][0]; return x; }il void work(){    n=gi(),m=gi(),Q=gi(); RG int top=0;    for (RG int i=1;i<=m;++i){ g[i].x=gi(),g[i].y=gi(),g[i].w=gi(); if (g[i].x>g[i].y) swap(g[i].x,g[i].y); }    sort(g+1,g+m+1,cmp); RG int k=0;    for (RG int i=1;i<=m;++i){ val[n+i]=g[i].w; if (g[i].x>g[i].y) swap(g[i].x,g[i].y); hsh[g[i].x][g[i].y]=i; }    for (RG int i=1;i<=Q;++i){	q[i].w=gi(),q[i].x=gi(),q[i].y=gi();	if (q[i].x>q[i].y) swap(q[i].x,q[i].y);	if (q[i].w==2) vis[hsh[q[i].x][q[i].y]]=1;    }    for (RG int i=1;i<=m;++i){	if (find(g[i].x)!=find(g[i].y) && !vis[hsh[g[i].x][g[i].y]]) link(g[i].x,n+i),link(g[i].y,n+i),k++;	if (k==n-1) break;    }    for (RG int i=Q;i;--i)	if (q[i].w==2){	    if (find(q[i].x)==find(q[i].y)){		RG int j=querymax(q[i].x,q[i].y); k=hsh[q[i].x][q[i].y];		if (val[j]>val[n+k]){		    cut(g[j-n].x,j),cut(g[j-n].y,j);		    link(q[i].x,n+k),link(q[i].y,n+k);		}	    }else link(q[i].x,n+i),link(q[i].y,n+i);	}else ans[++top]=val[querymax(q[i].x,q[i].y)];    for (RG int i=top;i;--i) PRintf("%d/n",ans[i]); return;}int main(){    File("pipe");    work();    return 0;}