LCT维护最小生成树，边权化点权。类似 P2387 [NOI2014]魔法森林（LCT）
离线存储询问，倒序处理，删边改加边。
#include<iostream> #include<cstdio> #include<cstring> #define rint register int using namespace std; inline void Swap(int &a,int &b){a^=b^=a^=b;} void read(int &x){ static char c=getchar();x=0; while(c<'0'||c>'9') c=getchar(); while('0'<=c&&c<='9') x=x*10+(c^48),c=getchar(); } #define M 200005 int n,m,Q,ch[M][2],fa[M],val[M],s[M],rev[M]; bool del[M]; int mp[1005][1005],p1[M],p2[M],p3[M],q1[M],q2[M],q3[M],ans[M]; #define lc ch[x][0] #define rc ch[x][1] inline bool nrt(int x){return ch[fa[x]][0]==x||ch[fa[x]][1]==x;} void up(int x){//维护最大边化为的点的编号 s[x]=x; if(val[s[lc]]>val[s[x]]) s[x]=s[lc]; if(val[s[rc]]>val[s[x]]) s[x]=s[rc]; } inline void Rev(int x){Swap(lc,rc),rev[x]^=1;} void down(int x){if(rev[x])Rev(lc),Rev(rc),rev[x]^=1;} void Pre(int x){if(nrt(x))Pre(fa[x]); down(x);} void turn(int x){ int y=fa[x],z=fa[y],l=(ch[y][1]==x),r=l^1; if(nrt(y)) ch[z][ch[z][1]==y]=x; fa[ch[x][r]]=y; fa[y]=x; fa[x]=z; ch[y][l]=ch[x][r]; ch[x][r]=y; up(y); up(x); } void splay(int x){ Pre(x); for(;nrt(x);turn(x)){ int y=fa[x],z=fa[y]; if(nrt(y)) turn((ch[y][1]==z)^(ch[z][1]==y)?x:y); } } void access(int x){for(int y=0;x;y=x,x=fa[x])splay(x),rc=y,up(x);} inline void makert(int x){access(x),splay(x),Rev(x);} int findrt(int x){ access(x);splay(x);down(x); while(lc) x=lc,down(x); splay(x); return x; } void link(int x,int y){makert(x); if(findrt(y)!=x) fa[x]=y;} void cut(int x,int y){ makert(x); if(findrt(y)==x&&fa[y]==x&&!ch[y][0]) fa[y]=rc=0,up(x); } inline void split(int x,int y){makert(x),access(y),splay(y);} void ins(int i){ bool is=1; if(findrt(p1[i])==findrt(p2[i])){//已经连在一起 split(p1[i],p2[i]); int w=s[p2[i]]; if(val[w]>p3[i]) cut(p1[w-n],w),cut(w,p2[w-n]);//删除最大边 else is=0; }if(is) link(p1[i],n+i),link(n+i,p2[i]); } int main(){ read(n);read(m);read(Q); for(rint i=1;i<=m;++i){ read(p1[i]),read(p2[i]),read(p3[i]); mp[p1[i]][p2[i]]=mp[p2[i]][p1[i]]=i; val[n+i]=p3[i]; } for(rint i=1;i<=Q;++i){ read(q1[i]),read(q2[i]),read(q3[i]); if(q1[i]==2) del[mp[q2[i]][q3[i]]]=1; } for(rint i=1;i<=m;++i) if(!del[i]) ins(i); for(rint i=Q;i>=1;--i){//倒序处理询问 if(q1[i]==2) ins(mp[q2[i]][q3[i]]); else split(q2[i],q3[i]),ans[i]=val[s[q3[i]]]; } for(rint i=1;i<=Q;++i) if(q1[i]==1) printf("%d\n",ans[i]); return 0; }