BZOJ4538 网络 [树链剖分]

那么对于每个询问我们只要判断权值大于K的路径的交是否都过x即可。路径的交还是路径，路径交满足结合律。可以离线拿个线段树维护一下即可，以权值为关键字，每个点记录该段区间的路径交
一道很标准的链交题，树链剖分+数据结构动态维护链上信息。转化信息时通过链交即可传递。
总之树链剖分练手。

#include<bits/stdc++.h>const int N = 101005;using namespace std;struct data{    int nxt,to;}E[N<<1];int head[N],sz[N],deep[N],dfn[N],dft[N],son[N],top[N],fa[N];bool vis[N<<1];int num,now,tot,ans,cnt,n,opt,m;inline void addedge(int x,int y){    E[++tot].nxt=head[x],head[x]=tot,E[tot].to=y;}void dfsf(int u,int f){    sz[u]=1,deep[u]=deep[f]+1;    for(register int v,i=head[u];i;i=E[i].nxt)        if(v=E[i].to,v!=f){            fa[v]=u,dfsf(v,u),sz[u]+=sz[v];            if(sz[son[u]]<=sz[v])son[u]=v;        }}void dfss(int u,int f){    top[u]=f,dft[dfn[u]=++cnt]=u;    if(son[u])dfss(son[u],f);    for(register int v,i=head[u];i;i=E[i].nxt)        if(v=E[i].to,v!=son[u]&&v!=fa[u])dfss(v,v);}struct node{    int x,y;    node(){}    node(int x,int y):x(x),y(y){}    inline bool operator < (const node &rhs)const{        return x<rhs.x;    }}a[N];priority_queue<node> q[N<<2];void insert(int x,int l,int r,int L,int R){    if(l==L&&r==R){        q[x].push(node(num,now));return ;    }    register int mid=(l+r)>>1;    if(R<=mid)insert(x<<1,l,mid,L,R);    else if(L>mid)insert(x<<1|1,mid+1,r,L,R);    else insert(x<<1,l,mid,L,mid),insert(x<<1|1,mid+1,r,mid+1,R);}void query(int x,int l,int r,int pos){    while(!q[x].empty()&&vis[q[x].top().y])q[x].pop();    if(!q[x].empty())ans=max(ans,q[x].top().x);    if(l==r)return;    register int mid=(l+r)>>1;    if(pos<=mid)query(x<<1,l,mid,pos);    else query(x<<1|1,mid+1,r,pos);}template<class T>inline void read(T &res){    static char ch;T flag=1;    while((ch=getchar())<'0'||ch>'9')if(ch=='-')flag=-1;res=ch-48;    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-48;res*=flag;}int main(){    read(n),read(m);    for(register int x,y,i=1;i<n;i++)        read(x),read(y),addedge(x,y),addedge(y,x);    dfsf(1,0),dfss(1,1);    for(now=1;now<=m;now++){        int x,y;        read(opt);        if(opt==0){            read(x),read(y),read(num);int flag=0;            for(;top[x]!=top[y];x=fa[top[x]]){                if(deep[top[x]]<deep[top[y]])swap(x,y);                a[++flag].x=dfn[top[x]],a[flag].y=dfn[x];            }            if(deep[x]>deep[y])swap(x,y);            a[++flag].x=dfn[x],a[flag].y=dfn[y];            sort(a+1,a+1+flag);            for(register int i=1;i<=flag;++i)                if(a[i-1].y+1<a[i].x)insert(1,1,n,a[i-1].y+1,a[i].x-1);            if(a[flag].y<n)insert(1,1,n,a[flag].y+1,n);        }else if(opt==1)read(x),vis[x]=1;        else{            read(x),ans=-1,query(1,1,n,dfn[x]),printf("%d\n",ans);        }    }    return 0;}