zoj2112 Dynamic Rankings （主席树 || 树套树）

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

题意：给你一个长为n的区间，有m个询问和修改，每次把下标i对应的值改变，或者询问某段区间的第k小值。

思路：这题是很经典的题，可以用主席树和树套树做，对于树套树的做法，我是用线段树的每一个节点都存一个treap，然后每一次修改就是单点更新，修改的时间复杂度为O(n*logn*logn)，询问时，需要二分值val，并统计线段树中对应的区间小于val值的个数，询问的时间复杂度为O(n*logn*logn*logn)，所以总的时间复杂度为O(n*logn*logn*logn)，同时因为线段树最多有logn层，每层都是ｎ个节点，所以treap的空间复杂度为O(n*logn)．

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
#define lson th<<1
#define rson th<<1|1
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 50005
int rt[maxn*4],a[maxn],n;
int cnt;
struct Treap{
    int key,pri,siz,num,son[2];
    void newnode(int x,int y){
        key=x;pri=y;
        siz=num=1;
        son[0]=son[1]=0;
    }
}T[20*maxn];

void rotate(int p,int &x)
{
    int y=T[x].son[!p];
    T[x].siz=T[x].siz-T[y].siz+T[T[y].son[p] ].siz;
    T[x].son[!p]=T[y].son[p];
    T[y].siz=T[y].siz-T[T[y].son[p] ].siz+T[x].siz;
    T[y].son[p]=x;
    x=y;
}

void charu(int key,int &x)
{
    if(x==0){
        x=++cnt;
        T[x].newnode(key,rand());
    }
    else{
        T[x].siz++;
        if(T[x].key==key){
            T[x].num++;return;
        }
        int p=key<T[x].key;
        charu(key,T[x].son[!p]);
        if(T[x].pri<T[T[x].son[!p] ].pri )
            rotate(p,x);
    }
}
void del(int key, int &x)
{
    if(T[x].key == key)
    {
        if(T[x].num>1){
            T[x].siz--;
            T[x].num--;
            return;
        }
        if(T[x].son[0] && T[x].son[1])
        {
            int p=T[T[x].son[0]].pri>T[T[x].son[1]].pri;
            rotate(p,x);
            del(key,x);
        }
        else x=T[x].son[1]+T[x].son[0];
    }
    else
    {
        T[x].siz--;
        int p=T[x].key>key;
        del(key,T[x].son[!p]);
    }
}
int query_rank(int key,int &x)
{
    if(x==0)return 0;
    if(T[x].key==key)return T[T[x].son[0] ].siz;
    if(T[x].key>key)return query_rank(key,T[x].son[0]);
    if(T[x].key<key)return T[T[x].son[0] ].siz+T[x].num+query_rank(key,T[x].son[1]);
}

void update(int idx,int val,int L,int R,int th,int f)
{
    if(f)del(a[idx],rt[th]);
    charu(val,rt[th]);
    if(L==idx && R==idx){
        a[idx]=val;
        return;
    }
    int mid=(L+R)/2;
    if(idx<=mid)update(idx,val,L,mid,lson,f);
    else update(idx,val,mid+1,R,rson,f);
}

int question(int l,int r,int val,int L,int R,int th)
{
    int mid;
    if(l==L && r==R){
        return query_rank(val,rt[th]);
    }
    mid=(L+R)/2;
    if(r<=mid)return question(l,r,val,L,mid,lson);
    else if(l>mid)return question(l,r,val,mid+1,R,rson);
    else return question(l,mid,val,L,mid,lson)+question(mid+1,r,val,mid+1,R,rson);
}

int main()
{
    int m,i,j,Tcase,l,r,c,d,e,mid;
    char s[10];
    scanf("%d",&Tcase);
    while(Tcase--)
    {
        memset(rt,0,sizeof(rt));
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        cnt=0;
        for(i=1;i<=n;i++){
            update(i,a[i],1,n,1,0);
        }
        for(i=1;i<=m;i++){
            scanf("%s",s);
            if(s[0]=='C'){
                scanf("%d%d",&c,&d);
                update(c,d,1,n,1,1);
            }
            else{
                scanf("%d%d%d",&c,&d,&e);
                l=0;r=1000000000;
                while(l<=r){
                    mid=(l+r)/2;
                    //printf("%d %d %d\n",l,r,mid);
                    if(question(c,d,mid,1,n,1)>=e)r=mid-1;
                    else l=mid+1;
                }
                printf("%d\n",r);
            }
        }

    }
    return 0;
}

/*
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
*/

主席树做法：

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define lth th<<1
#define rth th<<1|1
#define inf 99999999
#define pi acos(-1.0)
#define maxn 60005
#define M 2500010
int a[maxn],n;
struct node{
    int l,r,kind,d;
}ques[10005];
int pos[maxn],T[maxn],S[maxn];
int lson[M],rson[M],c[M];
int th,tot;

int build(int l,int r)
{
    int newroot=++th,i,j,mid;
    c[newroot]=0;   //!!!
    if(l!=r){
        mid=(l+r)/2;
        lson[newroot]=build(l,mid);
        rson[newroot]=build(mid+1,r);
    }
    return newroot;
}

int update(int root,int zhi,int value)
{
    int i,j;
    int newroot=++th;int tmp=newroot;
    int l=1,r=tot,mid;
    c[newroot]=c[root]+value;
    while(l<r){
        mid=(l+r)/2;
        if(zhi<=mid){
            r=mid;
            lson[newroot ]=++th;rson[newroot]=rson[root];
            newroot=lson[newroot];root=lson[root];
        }
        else{
            l=mid+1;
            lson[newroot ]=lson[root];rson[newroot]=++th;
            newroot=rson[newroot];root=rson[root];
        }
        c[newroot]=c[root]+value;
    }
    return tmp;
}

int lowbit(int x)
{
    return x&(-x);
}

void modify(int pos,int zhi,int value)
{
    int i,j;
    while(pos<=n){
        S[pos]=update(S[pos],zhi,value);
        pos+=lowbit(pos);
    }
}
int use[maxn];
int getsum(int pos)
{
    int sum=0;
    while(pos>0){
        sum+=c[lson[use[pos] ] ];
        pos-=lowbit(pos);
    }
    return sum;

}
int question(int left,int right,int k)
{
    int i,j,root1,root2;
    int l=1,r=tot,mid;
    for(i=left-1;i>0;i-=lowbit(i))use[i]=S[i];
    for(i=right;i>0;i-=lowbit(i))use[i]=S[i];
    root1=T[left-1];root2=T[right];

    while(l<r){
        mid=(l+r)/2;
        int tmp=getsum(right)-getsum(left-1)+c[lson[root2] ]-c[lson[root1] ];
        if(tmp>=k){
            r=mid;
            root1=lson[root1];root2=lson[root2];
            for(i=left-1;i>0;i-=lowbit(i))use[i]=lson[use[i] ];
            for(i=right;i>0;i-=lowbit(i))use[i]=lson[use[i] ];
        }
        else{
            k-=tmp;
            l=mid+1;
            root1=rson[root1];root2=rson[root2];
            for(i=left-1;i>0;i-=lowbit(i))use[i]=rson[use[i] ];
            for(i=right;i>0;i-=lowbit(i))use[i]=rson[use[i] ];
        }
    }
    return l;
}

int main()
{
    int m,i,j,Tcase,c,d,e;
    char str[5];
    scanf("%d",&Tcase);
    while(Tcase--)
    {
        scanf("%d%d",&n,&m);
        tot=0;
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
            tot++;
            pos[tot]=a[i];
        }
        for(i=1;i<=m;i++){
            scanf("%s",str);
            if(str[0]=='Q'){
                scanf("%d%d%d",&c,&d,&e);
                ques[i].kind=0;ques[i].l=c;ques[i].r=d;ques[i].d=e;
            }
            else if(str[0]=='C'){
                scanf("%d%d",&c,&d);
                tot++;pos[tot]=d;
                ques[i].kind=1;ques[i].l=c;ques[i].r=d;
            }
        }

        sort(pos+1,pos+1+tot);
        tot=unique(pos+1,pos+1+tot)-pos-1;

        th=0;
        T[0]=build(1,tot);

        for(i=1;i<=n;i++){
            int t=lower_bound(pos+1,pos+1+tot,a[i])-pos;
            T[i]=update(T[i-1],t,1);
        }

        for(i=1;i<=n;i++){
            S[i]=T[0];
        }
        for(i=1;i<=m;i++){
            if(ques[i].kind==0){   //表示询问
                printf("%d\n",pos[question(ques[i].l,ques[i].r,ques[i].d)]);
            }
            else{
                int t1=lower_bound(pos+1,pos+1+tot,a[ques[i].l])-pos;
                int t2=lower_bound(pos+1,pos+1+tot,ques[i].r)-pos;
                modify(ques[i].l,t1,-1);
                modify(ques[i].l,t2,1);
                a[ques[i].l ]=ques[i].r;  //!!!
            }
        }

    }
    return 0;
}

/*
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
*/