????在高中的時候就聽到過主席樹了，感覺非常高端，在寒假的時候 winter homework中有一題是查找區間第K大的樹，當時就開始百度這種網上的博客，發現主席樹看不懂，因為那個root[i]，還有tx[x].l與tx[x].r是什么意思沒有搞懂，所以那時候那題用了劃分樹的思想，最近有幸看到一篇博客，然后博主推薦了一個up主的視屏講解，最后終于弄懂了。

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K-th Number

Time Limit:?20000MS	?	Memory Limit:?65536K
Total Submissions:?64247	?	Accepted:?22601
Case Time Limit:?2000MS

Description

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.?
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"?
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).?
The second line contains n different integer numbers not exceeding 10⁹?by their absolute values --- the array for which the answers should be given.?
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output

For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

Sample Input

7 3 1 5 2 6 3 7 4 2 5 3 4 4 1 1 7 3

Sample Output

5 6 3

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

Source

Northeastern Europe 2004, Northern Subregion

? ?劃分樹模板

#include<cstdio> #include<cstring> #include<algorithm> #define MAXN 100010 using namespace std; int tree[20][MAXN],b[MAXN],p[20][MAXN]; void build(int l,int r,int dep) {if(l==r) return;int mid=(l+r)>>1,same=mid-l+1;for(int i=l;i<=r;i++)if(tree[dep][i]<b[mid]) same--;int lpos=l,rpos=mid+1;for(int i=l;i<=r;i++){if(tree[dep][i]<b[mid]) tree[dep+1][lpos++]=tree[dep][i];else if(tree[dep][i]==b[mid]&&same>0) {tree[dep+1][lpos++]=tree[dep][i];same--;}else tree[dep+1][rpos++]=tree[dep][i];p[dep][i]=p[dep][l-1]+lpos-l;} build(l,mid,dep+1);build(mid+1,r,dep+1); } int query(int L,int R,int l,int r,int dep,int k) {if(l==r) return tree[dep][l];int mid=(L+R)>>1;int cnt=p[dep][r]-p[dep][l-1];if(cnt>=k){int newl=L+p[dep][l-1]-p[dep][L-1],newr=newl+cnt-1;return query(L,mid,newl,newr,dep+1,k);}else{int newr=r+p[dep][R]-p[dep][r],newl=newr-(r-l-cnt);return query(mid+1,R,newl,newr,dep+1,k-cnt);} } int main(){int n,m;scanf("%d%d",&n,&m);for (int i = 1 ; i <= n; ++i){scanf("%d",&tree[0][i]);b[i] = tree[0][i];}sort(b+1,b+1+n);build(1,n,0);while(m--){int nl,nr,k;scanf("%d%d%d",&nl,&nr,&k);printf("%d\n",query(1,n,nl,nr,0,k));}return 0; }

主席樹模板

#include<cstdio> #include<iostream> #include<cstring> #include<queue> #include<cmath> #include<algorithm> using namespace std; const int maxn=1e5+6; int n,m,cnt,root[maxn],a[maxn],x,y,k;struct node{int l,r,sum; }T[maxn*40];vector<int> v; int getid(int x) {return lower_bound(v.begin(),v.end(),x)-v.begin()+1; }//離散化處理void update(int l,int r,int &x,int y,int pos) {T[++cnt]=T[y],T[cnt].sum++,x=cnt;if(l==r) return;int mid=(l+r)/2;if(mid>=pos) update(l,mid,T[x].l,T[y].l,pos);else update(mid+1,r,T[x].r,T[y].r,pos); } int query(int l,int r,int x,int y,int k) {if(l==r) return l;int mid=(l+r)/2;int sum=T[T[y].l].sum-T[T[x].l].sum;//左區間中數的個數是否大于k，if(sum>=k) return query(l,mid,T[x].l,T[y].l,k);else return query(mid+1,r,T[x].r,T[y].r,k-sum); }int main() {scanf("%d%d",&n,&m);for(int i=1;i<=n;i++) scanf("%d",&a[i]),v.push_back(a[i]);sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end());for(int i=1;i<=n;i++) update(1,n,root[i],root[i-1],getid(a[i]));for(int i=1;i<=m;i++){scanf("%d%d%d",&x,&y,&k);printf("%d\n",v[query(1,n,root[x-1],root[y],k)-1]); //前綴和的思想（y,x-1）} }

可以先看一下up主連接：主席樹?

root[i]表示1-i所形成的的線段樹起始的在T【】中的起始位置

T【x】.l表示左子樹的位置

T【x】.r表示右子樹的位置

通過l，r，pos的關系，劃分是左邊還是右邊（及它是第幾大的。和劃分樹的思想類似）

轉載于:https://www.cnblogs.com/The-Pines-of-Star/p/9878836.html

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