bzoj1146

这道题和bzoj2588很像，是动态区间第K大的变形。

先求DFS序，一棵子树的DFS是连续的，不妨记为[l,r]，我们维护前缀和，在l处+1，在r+1处-1。

变成动态区间第K大的经典问题，用树状数组套线段树。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj
 
using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define re(i,a,b)  for(i=a;i<=b;i++)
#define red(i,a,b) for(i=a;i>=b;i--)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

const DB EPS=1e-9;
inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);

inline int gint()
  {
        int res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }
inline LL gll()
  {
      LL res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }

const int maxN=80000;
const int maxQ=80000;
const int maxcnt=maxN+maxQ;

int N,Q;
int T[maxN+100];
int first[maxN+100],now;
struct Tedge{int v,next;}edge[2*maxN+100];
struct Tdata{int k,a,b;inline void input(){k=gint();a=gint();b=gint();}}data[maxQ+100];
int bak[maxcnt+100],cnt;

inline void addedge(int u,int v){now++;edge[now].v=v;edge[now].next=first[u];first[u]=now;}

int g,idx[maxN+100],l[maxN+100],r[maxN+100];
int vis[maxN+100];
int top,sta[maxN+100],last[maxN+100];
int dep[maxN+100],fa[maxN+100],jump[31][maxN+100];
int head,tail,que[maxN+100];
inline void DFS()
  {
      int j;
      g=0;
      vis[sta[top=1]=1]=1;
      last[1]=first[1];
        idx[1]=++g;
        fa[1]=0;
        dep[1]=1;
        jump[0][1]=1;re(j,1,30)jump[j][1]=jump[j-1][jump[j-1][1]];
        while(top>=1)
          {
              int u=sta[top],&i=last[top],v;
              for(v=edge[i].v;i!=-1;i=edge[i].next,v=edge[i].v)if(!vis[v])
                {
                    vis[sta[++top]=v]=1;
                    last[top]=first[v];
                    idx[v]=++g;
                    fa[v]=u;
                    dep[v]=dep[u]+1;
                    jump[0][v]=u;re(j,1,30)jump[j][v]=jump[j-1][jump[j-1][v]];
                    break;
                }
              if(i==-1)top--;
          }
        mmst(vis,0);
        vis[que[head=tail=0]=1]=1;
        while(head<=tail)
          {
              int u=que[head++],i,v;
              for(i=first[u],v=edge[i].v;i!=-1;i=edge[i].next,v=edge[i].v)if(!vis[v])vis[que[++tail]=v]=1;
          }
        red(j,tail,0)
          {
              int u=que[j],i,v;
              l[u]=r[u]=idx[u];
              for(i=first[u],v=edge[i].v;i!=-1;i=edge[i].next,v=edge[i].v)if(v!=jump[0][u])upmin(l[u],l[v]),upmax(r[u],r[v]);
          }
    }
            
struct Tnode{int son[2],val;}sn[12000000];int cn;
inline int newnode(){++cn;sn[cn].son[0]=sn[cn].son[1]=sn[cn].val=0;return cn;}
inline void update(int p,int l,int r,int x,int val)
  {
      while(l<=r)
        {
            sn[p].val+=val;
            if(l==r)break;
            int mid=(l+r)/2,f=(x>mid);
            if(!sn[p].son[f])sn[p].son[f]=newnode();
            p=sn[p].son[f];
            if(x<=mid)r=mid; else l=mid+1;
        }
    }

int tree[maxN+100];
#define lowbit(a) (a&(-a))
inline void change(int a,int x,int val)
  {
      for(;a<=g;a+=lowbit(a))
          update(tree[a],1,cnt,x,val);
  }
int r1,r2,r3,r4;
int t1[maxN+100],t2[maxN+100],t3[maxN+100],t4[maxN+100];
inline int ask(int p1,int p2,int p3,int p4,int k)
  {
      int i,l=1,r=cnt;
      r1=0;for(;p1>=1;p1-=lowbit(p1))t1[++r1]=tree[p1];
      r2=0;for(;p2>=1;p2-=lowbit(p2))t2[++r2]=tree[p2];
      r3=0;for(;p3>=1;p3-=lowbit(p3))t3[++r3]=tree[p3];
      r4=0;for(;p4>=1;p4-=lowbit(p4))t4[++r4]=tree[p4];
      while(l<=r)
        {
            if(l==r)return bak[l];
            int mid=(l+r)/2,as=0;
            re(i,1,r1)as+=sn[sn[t1[i]].son[1]].val;
            re(i,1,r2)as+=sn[sn[t2[i]].son[1]].val;
            re(i,1,r3)as-=sn[sn[t3[i]].son[1]].val;
            re(i,1,r4)as-=sn[sn[t4[i]].son[1]].val;
            if(k<=as)
              {
                  l=mid+1;
                  re(i,1,r1)t1[i]=sn[t1[i]].son[1];
                re(i,1,r2)t2[i]=sn[t2[i]].son[1];
                re(i,1,r3)t3[i]=sn[t3[i]].son[1];
                re(i,1,r4)t4[i]=sn[t4[i]].son[1];
              }
            else
              {
                  k-=as;
                  r=mid;
                  re(i,1,r1)t1[i]=sn[t1[i]].son[0];
                re(i,1,r2)t2[i]=sn[t2[i]].son[0];
                re(i,1,r3)t3[i]=sn[t3[i]].son[0];
                re(i,1,r4)t4[i]=sn[t4[i]].son[0];
              }
        }
  }

inline void swim(int &x,int H){int i;for(i=0;H!=0;H>>=1,i++)if(H&1)x=jump[i][x];}
inline int ask_lca(int x,int y)
  {
      if(dep[x]<dep[y])swap(x,y);
      swim(x,dep[x]-dep[y]);
      if(x==y)return x;
      int i;
      red(i,30,0)if(jump[i][x]!=jump[i][y]){x=jump[i][x];y=jump[i][y];}
      return jump[0][x];
  }

int main()
  {
      freopen("bzoj1146.in","r",stdin);
      freopen("bzoj1146.out","w",stdout);
      int i;
      N=gint();Q=gint();
      re(i,1,N)T[i]=gint();
      mmst(first,-1);now=-1;
      re(i,1,N-1){int x=gint(),y=gint();addedge(x,y);addedge(y,x);}
      re(i,1,Q)data[i].input();
      re(i,1,N)bak[++cnt]=T[i];
      re(i,1,Q)if(data[i].k==0)bak[++cnt]=data[i].b;
      sort(bak+1,bak+cnt+1);
      cnt=unique(bak+1,bak+cnt+1)-bak-1;
      re(i,1,N)T[i]=lower_bound(bak+1,bak+cnt+1,T[i])-bak;
      re(i,1,Q)if(data[i].k==0)data[i].b=lower_bound(bak+1,bak+cnt+1,data[i].b)-bak;
      DFS();
      re(i,1,g)tree[i]=newnode();
      re(i,1,N)
          {
              change(l[i],T[i],1);
                change(r[i]+1,T[i],-1);
            }
      re(i,1,Q)
        {
            int a=data[i].a,b=data[i].b,k=data[i].k;
            if(k==0)
              {
                  change(l[a],T[a],-1);change(r[a]+1,T[a],1);
                  T[a]=b;
                  change(l[a],T[a],1);change(r[a]+1,T[a],-1);
              }
            else
              {
                  int lca=ask_lca(a,b);
                  if(dep[a]+dep[b]-2*dep[lca]+1<k){printf("invalid request!
");continue;}
                  printf("%d
",ask(idx[a],idx[b],idx[lca],idx[fa[lca]],k));
              }
        }
      return 0;
  }