zoj2112

题目：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2112

经典的动态区间第K大。

用树状数组套线段树。

对原数组建一个树状数组，每个树状数组的结点代表一个线段树，这个线段树以权值为下标，包括这个树状数组的结点包含的区间。

插入的时候可以由树状数组和线段树写法类比，只与logn棵线段树有关，每棵线段树用时logn，总共的时间复杂度n*logn^2。

询问[l,r]的时候，类似与前缀和，找到与l-1有关的logn棵线段树，找到与人有关的logn棵线段树，然后类似于静态区间第K大，判断是往左子树还是右子树，有2logn棵线段树，判断往左子树还是右子树logn次，总共的时间复杂度n*logn^2。

网址上空间比较小，这个程序会MLE。

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

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

const int maxN=;
const int maxM=;
const int maxcnt=maxN+maxM;

int N,M;
int a[maxN+];
struct Tdata
  {
      char type;
      int l,r,k,t;
      inline void input()
        {
            type=getchar();while(!(type=='C' || type=='Q'))type=getchar();
            if(type=='Q'){l=gint();r=gint();k=gint();}else{l=gint();t=gint();}
        }
    }data[maxM+];

int bak[maxcnt+],cnt;

struct Tnode{int son[],val;}sn[maxcnt*+];int idx;
int tree[maxN+];

inline int newnode(){++idx;sn[idx].son[]=sn[idx].son[]=sn[idx].val=;return idx;}
inline void update(int p,int l,int r,int x,int val)
  {
      while()
        {
            sn[p].val+=val;
            if(l==r)break;
            int mid=(l+r)/;
            int f=(x>mid);
            if(!sn[p].son[f])sn[p].son[f]=newnode();
            if(x<=mid){p=sn[p].son[];r=mid;}else{p=sn[p].son[];l=mid+;}
        }
  }

#define lowbit(a) (a&(-a))
inline void change(int a,int x,int val)
  {
      for(;a<=N;a+=lowbit(a))
          update(tree[a],,cnt,x,val);
    }

int lge,larr[maxN+],rge,rarr[maxN+];
inline int ask(int l,int r,int k)
  {
      int i;
      l--;
      lge=;
      for(;l>=;l-=lowbit(l))larr[++lge]=tree[l];
      rge=;
      for(;r>=;r-=lowbit(r))rarr[++rge]=tree[r];
      int x=,y=cnt;
      while()
        {
            if(x==y)return bak[x];
            int mid=(x+y)/,G=;
            re(i,,rge)G+=sn[sn[rarr[i]].son[]].val;
            re(i,,lge)G-=sn[sn[larr[i]].son[]].val;
            if(G<k)
              {
                  k-=G;
                  x=mid+;
                  re(i,,rge)rarr[i]=sn[rarr[i]].son[];
                  re(i,,lge)larr[i]=sn[larr[i]].son[];
              }
            else
              {
                  y=mid;
                  re(i,,rge)rarr[i]=sn[rarr[i]].son[];
                  re(i,,lge)larr[i]=sn[larr[i]].son[];
              }
        }
  }

int main()
  {
      freopen("zoj2112.in","r",stdin);
      freopen("zoj2112.out","w",stdout);
      int i;
      for(int Case=gint();Case;Case--)
        {
            N=gint();M=gint();
            re(i,,N)a[i]=gint();
            re(i,,M)data[i].input();
            cnt=;
            re(i,,N)bak[++cnt]=a[i];
            re(i,,M)if(data[i].type=='C')bak[++cnt]=data[i].t;
            sort(bak+,bak+cnt+);
            cnt=unique(bak+,bak+cnt+)-bak-;
            re(i,,N)a[i]=lower_bound(bak+,bak+cnt+,a[i])-bak;
            re(i,,M)if(data[i].type=='C')data[i].t=lower_bound(bak+,bak+cnt+,data[i].t)-bak;
            idx=;
            re(i,,N)tree[i]=newnode();
            re(i,,N)change(i,a[i],);
            re(i,,M)
              {
                  int x;
                  switch(data[i].type)
                    {
                        case 'C':
                            x=data[i].l;
                            change(x,a[x],-);
                            a[x]=data[i].t;
                            change(x,a[x],);
                        break;
                        case 'Q':
                            printf("%d\n",ask(data[i].l,data[i].r,data[i].k));
                        break;
                    }
              }
        }
      return ;
  }