ZOJ 2112 Dynamic Rankings(树状数组+主席树)

Posted 2022-06-22 songorz

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.

 

<b< dd="">

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.

 

<b< dd="">

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

 题解：

树状数组的每个节点都是一颗线段树，但这棵线段树不再保存每个前缀的信息了，而是由树状数组的sum函数计算出这个前缀的信息，那么显而易见这棵线段树保存的是辅助数组S的值，即S=A[i-lowbit+1]+...+A[i],其中A[i]表示值为i的元素出现的次数。

那么对于每次修改，我们要修改树状数组上的logn棵树，对于每棵树，我们要修改logn个结点，所以时空复杂度为

O(（n+q）*logn*logn)，由于这道题n比较大，查询次数q比较小，所以我们可以初始时建立一颗静态的主席树，树状数组只保存每次修改的信息，那么时空复杂度降为了O（n*logn+q*logn*logn）
参考代码：

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) (x&-x)
typedef long long ll;
inline int read()

    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘)if(ch==‘-‘) f=-1;ch=getchar();
    while(ch>=‘0‘&&ch<=‘9‘)x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();
    return x*f;

const int maxn=6e4+10;
struct Tree
    int ls,rs;
    int sum;
 node[maxn*34];

struct Qnode
    bool flag;
    int l,r,k;
 qn[maxn];

int T,n,q,a[maxn],b[maxn],num,rt[maxn];
char str[2];
int ul[maxn],ur[maxn],ca[maxn],s[maxn],cnt;

void Build(int rt,int l,int r)

    rt=++cnt;
    node[rt].sum=0;
    if(l==r) return ;
    int mid=l+r>>1;
    Build(node[rt].ls,l,mid);
    Build(node[rt].rs,mid+1,r);


void Update(int y,int &x,int l,int r,int pos,int val)

    node[++cnt]=node[y];node[cnt].sum+=val;x=cnt;
    if(l==r) return ;
    int mid=l+r>>1;
    if(pos<=mid) Update(node[y].ls,node[x].ls,l,mid,pos,val);
    else Update(node[y].rs,node[x].rs,mid+1,r,pos,val);


void Add(int x,int val)

    int res=lower_bound(b+1,b+1+num,a[x])-b;
    while(x<=n)
    
        Update(s[x],s[x],1,num,res,val);
        x+=lowbit(x);
    


int Sum(int x,bool temp)

    int res=0;
    while(x>0)
    
        if(temp) res+=node[node[ur[x]].ls].sum;
        else res+=node[node[ul[x]].ls].sum;
        x-=lowbit(x);
    
    return res;


int Query(int L,int R,int x,int y,int l,int r,int k)

    if(l==r) return l;
    int mid=l+r>>1;
    int res=Sum(R,true)-Sum(L,false)+node[node[y].ls].sum-node[node[x].ls].sum;
    if(k<=res)
    
        for(int i=R;i;i-=lowbit(i)) ur[i]=node[ur[i]].ls;
        for(int i=L;i;i-=lowbit(i)) ul[i]=node[ul[i]].ls;
        return Query(L,R,node[x].ls,node[y].ls,l,mid,k);
    
    else
    
        for(int i=R;i;i-=lowbit(i)) ur[i]=node[ur[i]].rs;
        for(int i=L;i;i-=lowbit(i)) ul[i]=node[ul[i]].rs;
        return Query(L,R,node[x].rs,node[y].rs,mid+1,r,k-res);
    


int main()

    T=read();
    while(T--)
    
        n=read();q=read(); cnt=num=0;
        memset(rt,0,sizeof(rt));
        for(int i=1;i<=n;++i) a[i]=read(),b[++num]=a[i];
        for(int i=1;i<=q;++i)
        
            scanf("%s",str);
            if(str[0]==‘Q‘)
            
                qn[i].flag=true;
                qn[i].l=read();qn[i].r=read();qn[i].k=read();
            
            else
            
                qn[i].flag=false;
                qn[i].l=read();qn[i].r=read();b[++num]=qn[i].r;
            
        
        sort(b+1,b+1+num);
        int tot=unique(b+1,b+1+num)-b-1;
        num=tot;
        for(int i=1;i<=n;++i) ca[i]=lower_bound(b+1,b+1+num,a[i])-b;
        Build(rt[0],1,num);
        for(int i=1;i<=n;++i) Update(rt[i-1],rt[i],1,num,ca[i],1);
        for(int i=1;i<=n;++i) s[i]=rt[0];
        for(int i=1;i<=q;++i)
        
            if(qn[i].flag)
            
                for(int j=qn[i].r;j;j-=lowbit(j)) ur[j]=s[j];
                for(int j=qn[i].l-1;j;j-=lowbit(j)) ul[j]=s[j];
                printf("%d\n",b[Query(qn[i].l-1,qn[i].r,rt[qn[i].l-1],rt[qn[i].r],1,num,qn[i].k)]);
            
            else
            
                Add(qn[i].l,-1);
                a[qn[i].l]=qn[i].r;
                Add(qn[i].l,1);
            
        
    


    return 0;

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