[Knowledge Learning] Chairman tree

Two days learning the Chairman of the tree, the tree is basically get to know how to operate the chairman of

Chairman of the tree, is a kind of persistent segment tree. The simplest operation is to maintain a static Interval \ (k \) small

Chairman of the tree through the relevant operation and maintenance history versions for query interval

Chairman of the principle of the tree

Suppose now that there is such a sequence: \ (4,1,3,5,2 \)

Asked how determined interval [1,3] in the size of the second number?

~~With a large eye observation, it is clear that 3~~

So let the computer to how to achieve it?~~He has no eyes~~

For this sequence, we can start to build an empty weight segment tree, named "Tree 0" (easy to use back), as shown:

Do not tell me you do not know what is the right value segment tree of their own to Baidu;

Now there first sequence number is \ (4 \) , we insert into the tree inside a \ (4 \) , because you want to keep the version history, so we \ (4 \) Create a segment tree this number, name "tree 1", as shown:

Why is this so?

\ (. 1 \ Le. 4 \ LE5 \) , so that the interval [1,5] ++;

\ (. 4 \ Le. 4 \ LE5 \) , so that the interval [4,5] ++;

\ (. 4 \ Le. 4 \ LE4 \) , so that the interval [4,4] ++;

Other or \ (0 \) ;

Do not you see. . .

Further inserted second number \ (1 \) , into a "tree 2", not explained here

And then inserted into the third number \ (3 \) , into a "tree 3"

OK! Now we can already find the size of [1,3] is the second largest in the number of

Recursive queries will be ranked right?

Do not look at this:

进入[1，5]节点，我们发现他的左儿子的子树个数为$2$ ， $2\le k $ $(k=2)$，于是进入[1，3]节点;
然后我们发现[1，3]节点的左儿子子树个数$1 < k$ $(k=2)$，于是进入[3，3]节点；
此时我们把$k$更新为$1$ ($2-1=1$);
走到头了，于是就返回3，所以答案就是3，也就是原来的序列区间[1, 3]的第2小就是3

现在你明白了主席树是怎么操作了的吧？

疑问

但是有一个问题：上面我们求的是区间[1，3]的第 $k$ 大的数

同理，区间[1，r] ($r \in [1，n]，r \in N$)的第 $k$ 大数我们也就会求了

那怎么求区间[l，r]的第 $k$ 大数呢？

举个例子，求区间[2，3]的第 $k$ 大数

我们拿建出来的“树3”减掉“树1”后，再进行如上操作就可以了

这也就是前缀和思想

所以对于区间[l，r] 我们拿“树r”减去“树(l-1)”，再query一下就可以求得答案了

例题与代码

那么主席树就介绍完了，具体实现给个例题让大家看看，还有不懂得可以再参考一下他人的博客

例题：静态区间第 $k$ 小（【模板】可持久化线段树1/主席树）

题目链接

（鸣谢@wsk1202提供板子）

#include <bits/stdc++.h>
#define N (200000+5)
#define ls ch[rt][0]
#define rs ch[rt][1]
#define vl ch[vs][0]
#define vr ch[vs][1]//几个宏定义
using namespace std;
int n,m,q;
int rt[N],ch[N<<5][2],tot,val[N<<5];
int a[N],b[N];
inline int query(int x){//离散化
    return lower_bound(b+1,b+m+1,x)-b;
}
inline void update(int &rt,int vs,int l,int r,int k){//新建一棵树
    rt=++tot;
    ls=vl,rs=vr;
    val[rt]=val[vs]+1;
    if(l==r) return;
    int mid=(l+r)>>1;
    if(k<=mid) update(ls,vl,l,mid,k);
    else update(rs,vr,mid+1,r,k);
}
inline int query(int rt,int vs,int l,int r,int k){//询问
    if(l==r) return l;
    int mid=(l+r)>>1;
    int v=val[vl]-val[ls];
    if(k<=v) return query(ls,vl,l,mid,k);
    else return query(rs,vr,mid+1,r,k-v);
}
int main(){
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]),b[i]=a[i];
    sort(b+1,b+n+1);
    m=unique(b+1,b+n+1)-b-1;
    for(int i=1;i<=n;i++){
        update(rt[i],rt[i-1],1,n,query(a[i]));
    }
    while(q--){
        int l,r,k;
        scanf("%d%d%d",&l,&r,&k);
        printf("%d\n",b[query(rt[l-1],rt[r],1,n,k)]);
    }
    return 0;
}