【BZOJ 1146】【CTSC 2008】网络管理network

一句话题意，树链上带改动区间第k大
感觉能够dfs+主席树$O(n\log^2n)$过掉，但我不会写= =
于是写的线段树套平衡树+链剖+二分（改动$O(n\log^3n)$,查询$O(n\log^4n)$慢了好多啊QAQ）
这里简介一下区间第K大做法。对于每一个线段树所”管辖“的范围，建一棵相应范围内的平衡树（我用的Treap）；改动时，改动每一个包括被改动节点的线段树节点所相应的Treap。查询时。二分
答案。统计每一个区间内比当前答案小的数就可以（为了保证是序列里的数。我们能够二分答案在原序列中排名）
PS：这题真的是第K大，不是排名第K的，被坑WA了一次= =
code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define mid (l+r)/2
#define lch i<<1,l,mid
#define rch i<<1|1,mid+1,r
using namespace std;
struct treap_node{
    treap_node *left,*right;
    int val,fix,size,wgt;
    treap_node(int val): val(val) {size=1; wgt=1; left=right=NULL; fix=rand();}
    int lsize()
      {if (left) return left->size; else return 0;}
    int rsize()
      {if (right) return right->size; else return 0;}
    void Maintain()
      {size=wgt; size+=lsize()+rsize();}
};
treap_node *seg[320001];
int f[80001],plc[80001];
int point[80001],next[200001];
struct hp{
    int u,v;
}ai[200001];
struct hq{
    int dep,fat,top,size,wson;
}tree[80001];
int n,a[80001],ans,m,totw,e=0;
void tlr(treap_node *&a)
{
    treap_node *b=a->right;
    a->right=b->left; b->left=a;
    a->Maintain(); b->Maintain(); a=b;
}
void trr(treap_node *&a)
{
    treap_node *b=a->left;
    a->left=b->right; b->right=a;
    a->Maintain(); b->Maintain(); a=b;
}
void add(int u,int v)
{
    e++; ai[e].u=u; ai[e].v=v; next[e]=point[u]; point[u]=e;
    e++; ai[e].v=u; ai[e].v=u; next[e]=point[v]; point[v]=e;
}
void insert(treap_node *&p,int value)  
{  
    if (!p)  
      p=new treap_node(value);  
    else  
      {  
        if (value==p->val)  
          p->wgt++;  
        if (value<p->val)  
          {  
            insert(p->left,value);  
            if (p->left->fix<p->fix)  
              trr(p);  
          }  
        if (value>p->val)  
          {  
            insert(p->right,value);  
            if (p->right->fix<p->fix)  
              tlr(p);  
          }  
      }  
    p->Maintain();  
}
void make_node(int i,int l,int r)
{
    int j;
    for (j=l;j<=r;++j)
      insert(seg[i],a[f[j]]);
}
void build(int i,int l,int r)
{
    make_node(i,l,r);
    if (l==r) return;
    build(lch); build(rch);
}
void build_tree(int now,int last,int depth)
{
    int i;
    tree[now].fat=last;
    tree[now].dep=depth;
    tree[now].size=1;
    tree[now].wson=0;
    for (i=point[now];i;i=next[i])
      if (ai[i].v!=last)
        {
          build_tree(ai[i].v,now,depth+1);
          tree[now].size+=tree[ai[i].v].size;
          if (tree[ai[i].v].size>tree[tree[now].wson].size)
            tree[now].wson=ai[i].v;
        }
}
void build_seg(int now,int tp)
{
    int i;
    tree[now].top=tp;
    plc[now]=++totw; f[totw]=now;
    if (tree[now].wson!=0)
      build_seg(tree[now].wson,tp);
    for (i=point[now];i;i=next[i])
      if (ai[i].v!=tree[now].wson&&ai[i].v!=tree[now].fat)
        build_seg(ai[i].v,ai[i].v);
}
void del(treap_node *&p,int val)
{
    if (val==p->val)
      {
        if (p->wgt==1)
          {
            if (!p->left||!p->right)
              {
                if (!p->left) p=p->right;
                else p=p->left;
              }
            else
              {
                if (p->left->fix<p->right->fix)
                  {trr(p); del(p->right,val);}
                else
                  {tlr(p); del(p->left,val);}
              }
          }
        else
          p->wgt--;
      }
    else
      {
        if (val<p->val) del(p->left,val);
        else del(p->right,val);
      }
    if (p!=NULL) p->Maintain(); 
}
int kth(treap_node *p,int k)
{
    if (k<=p->lsize()) return kth(p->left,k);
    if (k>p->lsize()+p->wgt) return kth(p->right,k-p->lsize()-p->wgt);
    if (k<=p->lsize()+p->wgt) return p->val;
}
int rank(treap_node *p,int val,int cur)
{
    if (val==p->val) return cur+p->lsize();
    if (val>p->val&&!p->right) return cur+p->lsize()+p->wgt;
    if (val<p->val&&!p->left) return cur;
    if (val<p->val) return rank(p->left,val,cur);
    if (val>p->val) return rank(p->right,val,cur+p->lsize()+p->wgt);
}
void query(int i,int l,int r,int x,int y,int val)
{
    if (x<=l&&y>=r)
      {
        ans+=rank(seg[i],val,0);
        return;  
      }
    if (x<=mid) query(lch,x,y,val);
    if (y>mid) query(rch,x,y,val);
}
void delt(int i,int l,int r,int x,int num)
{
    if (l==x&&l==r)
      {
        del(seg[i],num);
        return;
      }
    del(seg[i],num);
    if (x<=mid) delt(lch,x,num);
    else delt(rch,x,num);
}
void ins(int i,int l,int r,int x,int num)
{
    if (l==x&&l==r)
      {
        insert(seg[i],num);
        return;
      }
    insert(seg[i],num);
    if (x<=mid) ins(lch,x,num);
    else ins(rch,x,num);
}
int Qsum(int x,int y)
{
    int t=0,f1=tree[x].top,f2=tree[y].top;
    while (f1!=f2)
      {
        //cout<<x<<‘ ‘<<y<<‘ ‘<<f1<<‘ ‘<<f2<<endl;
        if (tree[f1].dep<tree[f2].dep) {swap(x,y); swap(f1,f2);}
        t+=plc[x]-plc[f1]+1;
        x=tree[f1].fat; f1=tree[x].top;
      }
    if (tree[x].dep>tree[y].dep) swap(x,y);
    t+=plc[y]-plc[x]+1;
    return t;
}
void Q(int x,int y,int num)
{
    int f1=tree[x].top,f2=tree[y].top;
    while (f1!=f2)
      {
        if (tree[f1].dep<tree[f2].dep) {swap(x,y); swap(f1,f2);}
        query(1,1,n,plc[f1],plc[x],num);
        x=tree[f1].fat; f1=tree[x].top;
      } 
    if (tree[x].dep>tree[y].dep) swap(x,y);
    query(1,1,n,plc[x],plc[y],num);
}
void work(int x,int y,int k)
{
    int l,r,t,midx,len;
    l=1; r=n;
    len=Qsum(x,y);
    if (k>len) {printf("invalid request!\n"); return;}
    k=len-k+1;
    while (l<r)
      {
        midx=(l+r+1)/2;
        t=kth(seg[1],midx);
        ans=0; Q(x,y,t);
        if (ans<=k-1) l=midx;
        else r=midx-1;
      }
    printf("%d\n",kth(seg[1],l));
}
int main()
{
    int i,x,y,k;
    scanf("%d%d",&n,&m);
    for (i=1;i<=n;++i)
      scanf("%d",&a[i]);
    for (i=1;i<n;++i)
      {
        scanf("%d%d",&x,&y);
        add(x,y);
      }
    build_tree(1,0,0);
    build_seg(1,1);
    build(1,1,n);
    for (i=1;i<=m;++i)
      {
        scanf("%d%d%d",&k,&x,&y);
        if (k==0)
          {
            delt(1,1,n,plc[x],a[x]);
            ins(1,1,n,plc[x],y);
            a[x]=y; 
          }
        if (k>0)
          work(x,y,k);
      }
}