ZOJ2112-Dynamic Rankings（树状数组套主席树）

Dynamic Rankings

Time Limit: 10 Seconds Memory Limit: 32768 KB

The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1], a[2], …, a[N], you can ask it like: what is the k-th smallest number of a[i], a[i+1], …, a[j]? (For some i<=j, 0< k<=j+1-i that you have given to it). More powerful, you can even change the value of some a[i], and continue to query, all the same.

Your task is to write a program for this computer, which

• Reads N numbers from the input (1 <= N <= 50,000)

• Processes M instructions of the input (1 <= M <= 10,000). These instructions include querying the k-th smallest number of a[i], a[i+1], …, a[j] and change some a[i] to t.

Input

The first line of the input is a single number X (0 < X <= 4), the number of the test cases of the input. Then X blocks each represent a single test case.

The first line of each block contains two integers N and M, representing N numbers and M instruction. It is followed by N lines. The (i+1)-th line represents the number a[i]. Then M lines that is in the following format

Q i j k or
C i t

It represents to query the k-th number of a[i], a[i+1], …, a[j] and change some a[i] to t, respectively. It is guaranteed that at any time of the operation. Any number a[i] is a non-negative integer that is less than 1,000,000,000.

There’re NO breakline between two continuous test cases.

Output

For each querying operation, output one integer to represent the result. (i.e. the k-th smallest number of a[i], a[i+1],…, a[j])

There’re NO breakline between two continuous test cases.

Sample Input

2
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3

Sample Output

3
6
3
6

题目大意：动态第k大
解题思路：树状数组套主席树

#include<iostream>#include<cstdio>#include<string>#include<cmath>#include<vector>#include<map>#include<set>#include<queue>#include<algorithm>#include<cstring>using namespace std;typedef long long LL;const int INF=0x3f3f3f3f;const int MAXN=6e4+100;const int M=MAXN*40;int n,q,m,tot;int a[MAXN],t[MAXN];int T[MAXN],lson[M],rson[M],c[M];int S[MAXN];struct Query{  int kind;  int l,r,k;}query[10005];void inithash(int k){  sort(t,t+k);  m=unique(t,t+k)-t;}int Hash(int x){  return lower_bound(t,t+m,x)-t;}int build(int l,int r){  int root=tot++;  c[root]=0;  if(l!=r)  {    int mid=(l+r)/2;    lson[root]=build(l,mid);    rson[root]=build(mid+1,r);  }  return root;}int Insert(int root,int pos,int val){  int newroot=tot++,tmp=newroot;  int l=0,r=m-1;  c[newroot]=c[root]+val;  while(l<r)  {    int mid=(l+r)>>1;    if(pos<=mid)    {      lson[newroot]=tot++;      rson[newroot]=rson[root];      newroot=lson[newroot];      root=lson[root];      r=mid;    }else    {      rson[newroot]=tot++;      lson[newroot]=lson[root];      newroot=rson[newroot];      root=rson[root];      l=mid+1;    }    c[newroot]=c[root]+val;  }  return tmp;}int lowbit(int x){  return x&(-x);}int use[MAXN];void add(int x,int pos,int val){  while(x<=n)  {    S[x]=Insert(S[x],pos,val);    x+=lowbit(x);  }}int sum(int x){  int ret=0;  while(x>0)  {    ret+=c[lson[use[x]]];    x-=lowbit(x);  }  return ret;}int Query(int left,int right,int k){  int leftroot=T[left-1];  int rightroot=T[right];  int l=0,r=m-1;  for(int i=left-1;i;i-=lowbit(i)) use[i]=S[i];  for(int i=right;i;i-=lowbit(i))  use[i]=S[i];  while(l<r)  {    int mid=(l+r)/2;    int tmp=sum(right)-sum(left-1)+c[lson[rightroot]]-c[lson[leftroot]];    if(tmp>=k)    {      r=mid;      for(int i=left-1;i;i-=lowbit(i)) use[i]=lson[use[i]];      for(int i=right;i;i-=lowbit(i))  use[i]=lson[use[i]];      leftroot=lson[leftroot];      rightroot=lson[rightroot];    }else    {      l=mid+1;      k-=tmp;      for(int i=left-1;i;i-=lowbit(i)) use[i]=rson[use[i]];      for(int i=right;i;i-=lowbit(i)) use[i]=rson[use[i]];      leftroot=rson[leftroot];      rightroot=rson[rightroot];    }  }  return l;}void Modify(int x,int p,int d){  while(x<=n)  {    S[x]=Insert(S[x],p,d);    x+=lowbit(x);  }}int main(){  int cas;  scanf("%d",&cas );  while(cas--)  {    scanf("%d%d",&n,&q );    tot=0;    m=0;    for(int i=1;i<=n;i++)    {      scanf("%d",&a[i] );      t[m++]=a[i];    }    char op[10];    for(int i=0;i<q;i++)    {      scanf("%s",op);      if(op[0]=='Q')      {        query[i].kind=0;        scanf("%d%d%d",&query[i].l,&query[i].r,&query[i].k );      }else      {        query[i].kind=1;        scanf("%d%d",&query[i].l,&query[i].r );        t[m++]=query[i].r;      }    }    inithash(m);    T[0]=build(0,m-1);    for(int i=1;i<=n;i++)  T[i]=Insert(T[i-1],Hash(a[i]),1);    for(int i=1;i<=n;i++)  S[i]=T[0];    for(int i=0;i<q;i++)    {      if(query[i].kind==0)          printf("%d\n", t[Query(query[i].l,query[i].r,query[i].k)]);      else      {        Modify(query[i].l,Hash(a[query[i].l]),-1);        Modify(query[i].l,Hash(query[i].r),1);        a[query[i].l]=query[i].r;      }    }  }}/*25 33 2 1 4 7Q 1 4 3C 2 6Q 2 5 35 33 2 1 4 7Q 1 4 3C 2 6Q 2 5 33636*/