Query on a tree III

最新推荐文章于 2020-04-09 21:17:23 发布

转载最新推荐文章于 2020-04-09 21:17:23 发布 · 156 阅读

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原文链接：http://www.cnblogs.com/DexterYsw/p/7954997.html

本文介绍了一种在树状结构中进行高效查询的算法，主要针对查询子树中第k大的节点标签问题。通过使用主席树的方法，在深度优先搜索序列上进行操作，实现了对特定子树内节点标签的有效查询。

Query on a tree III

You are given a node-labeled rooted tree with n nodes.
Define the query (x, k): Find the node whose label is k-th largest in the subtree of the node x. Assume no two nodes have the same labels.
The first line contains one integer n (1 <= n <= 105). The next line contains n integers li (0 <= li <= 109) which denotes the label of the i-th node.
Each line of the following n - 1 lines contains two integers u, v. They denote there is an edge between node u and node v. Node 1 is the root of the tree.

The next line contains one integer m (1 <= m <= 104) which denotes the number of the queries. Each line of the next m contains two integers x, k. (k <= the total node number in the subtree of x)

For each query (x, k), output the index of the node whose label is the k-th largest in the subtree of the node x.

Input:

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

Output:

5
4
5
5

Solution

查询第k大就是主席树裸体
然后这个就是查询子树第k大
那不就是 dfs序上查询即可？

#include <bits/stdc++.h>
using namespace std;
#define maxn (int)(1e5+10)
#define LL long long

inline int read(){
    int rtn=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))rtn=(rtn<<1)+(rtn<<3)+ch-'0',ch=getchar();
    return rtn*f;
}

int n,m,cnt,p[maxn],w[maxn],dfn[maxn],root[maxn],size[maxn],coc,num,sz;

struct node{
    int a,b,nt;
}e[maxn<<2];

struct HJT{
    int ls,rs,cnt,id;
}T[maxn*60];

inline void add(int x,int y){
    e[++cnt].a=x;e[cnt].b=y;
    e[cnt].nt=p[x];p[x]=cnt;
}

inline void insert(int &p,int l,int r,int q,int v){
    T[++sz]=T[p];p=sz;
    if(l==r)return void((T[p].cnt=1)&&(T[p].id=v));
    int mid=l+r>>1;
    if(q<=mid)insert(T[p].ls,l,mid,q,v);
    else if(q>mid)insert(T[p].rs,mid+1,r,q,v);
    T[p].cnt=T[T[p].ls].cnt+T[T[p].rs].cnt;
}

inline int query(int lp,int rp,int l,int r,int k){
    if(l==r)return T[rp].id;
    int s=T[T[rp].ls].cnt-T[T[lp].ls].cnt;
    int mid=l+r>>1;
    if(k<=s)return query(T[lp].ls,T[rp].ls,l,mid,k);
    else return query(T[lp].rs,T[rp].rs,mid+1,r,k-s); 
}

inline void dfs(int x){
    size[x]=1;dfn[x]=++coc;
    root[coc]=root[coc-1];
    insert(root[coc],0,1e9,w[x],x);
    for(int i=p[x];i;i=e[i].nt){
        int k=e[i].b;
        if(size[k])continue;
        dfs(k);size[x]+=size[k];
    }
}

int main(){
    n=read();
    for(int i=1;i<=n;i++)scanf("%d",&w[i]);
    for(int i=1;i<n;i++){
        int x=read(),y=read();
        add(x,y);add(y,x);
    }
    dfs(1);
    m=read();
    while(m--){
        int x=read(),k=read();
        printf("%d\n",query(root[dfn[x]-1],root[dfn[x]+size[x]-1],0,1e9,k));
    }
    return 0;
}

转载于:https://www.cnblogs.com/DexterYsw/p/7954997.html