Query on A Tree HDU - 6191 （离线+01字典树合并）

最新推荐文章于 2021-11-09 18:22:44 发布

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本文介绍了一种利用字典树解决特定树形结构中节点值与给定整数进行异或运算的问题，通过构建字典树并实现合并、查询等操作，有效地解决了寻找最大异或结果的问题。

Monkey A lives on a tree, he always plays on this tree.

One day, monkey A learned about one of the bit-operations, xor. He was keen of this interesting operation and wanted to practise it at once.

Monkey A gave a value to each node on the tree. And he was curious about a problem.

The problem is how large the xor result of number x and one node value of label y can be, when giving you a non-negative integer x and a node label u indicates that node y is in the subtree whose root is u(y can be equal to u).

Can you help him?
Input
There are no more than 6 test cases.

For each test case there are two positive integers n and q, indicate that the tree has n nodes and you need to answer q queries.

Then two lines follow.

The first line contains n non-negative integers V1,V2,⋯,Vn, indicating the value of node i.

The second line contains n-1 non-negative integers F1,F2,⋯Fn−1, Fi means the father of node i+1.

And then q lines follow.

In the i-th line, there are two integers u and x, indicating that the node you pick should be in the subtree of u, and x has been described in the problem.

2≤n,q≤105

0≤Vi≤109

1≤Fi≤n, the root of the tree is node 1.

1≤u≤n,0≤x≤109
Output
For each query, just print an integer in a line indicating the largest result.
Sample Input
2 2
1 2
1
1 3
2 1
Sample Output
2
3

和线段树合并类似，而且好写很多，本质上其实都是黑箱操作，把两个数据集合并就是了，数据结构的使用其实降低了很多思维难度。
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#define maxx 100005
using namespace std;
int head[maxx];
int to[maxx],_next[maxx];
int val[maxx];
int cnt;
void addEdge(int u,int v)
{
    to[++cnt]=v,_next[cnt]=head[u],head[u]=cnt;
}

int n,q;
struct node
{
    int ind;
    int x;
    node(){}
    node(int _ind,int _x):ind(_ind),x(_x){}
};
vector<node>query[maxx];//询问关联的节点
struct Trie
{
    int val;
    Trie* next[2];
}*root[maxx];
void _insert(Trie* rt,int x)//字典树插入
{
    for(int i=30;i>=0;i--)
    {
        int id=(x>>i)&1;
        if(rt->next[id]==NULL)
        {
            rt->next[id]=new Trie;
            rt->next[id]->next[0]=NULL;
            rt->next[id]->next[1]=NULL;
        }
        rt=rt->next[id];
    }
    rt->val=x;
}
int _query(Trie* rt,int x)//字典树询问
{
    for(int i=30;i>=0;i--)
    {
        int id=(x>>i)&1;
        if(rt->next[id^1]!=NULL)
            rt=rt->next[id^1];
        else
            rt=rt->next[id];
    }
    return rt->val^x;
}
Trie* _merge(Trie* a,Trie* b)//字典树合并，非常好写
{
    if(a==NULL)return b;
    if(b==NULL)return a;
    a->next[0]=_merge(a->next[0],b->next[0]);
    a->next[1]=_merge(a->next[1],b->next[1]);
    delete(b);
    return a;
}
int ans[maxx];
void dfs(int u)
{
    root[u]=new Trie;
    root[u]->next[0]=root[u]->next[1]=NULL;
    _insert(root[u],val[u]);
    for(int i=head[u];i;i=_next[i])
    {
        int v=to[i];
        dfs(v);
        root[u]=_merge(root[u],root[v]);
    }
    for(int i=0;i<query[u].size();i++)
    {
        node qq=query[u][i];
        ans[qq.ind]=_query(root[u],qq.x);
    }
}
void clean(Trie* rt)
{
    if(rt->next[0]!=NULL)clean(rt->next[0]);
    if(rt->next[1]!=NULL)clean(rt->next[1]);
    delete(rt);
}
void init()
{
    cnt=0;
    memset(head,0,sizeof(head));
    for(int i=1;i<=n;i++)
        query[i].clear();
}
int main()
{
    while(scanf("%d%d",&n,&q)==2)
    {
        init();
        for(int i=1;i<=n;i++)
            scanf("%d",val+i);
        int u,x;
        for(int i=2;i<=n;i++)
            scanf("%d",&u),addEdge(u,i);
        for(int i=1;i<=q;i++)
        {
            scanf("%d%d",&u,&x);
            query[u].push_back(node(i,x));
        }
        dfs(1);
        for(int i=1;i<=q;i++)
            printf("%d\n",ans[i]);
        clean(root[1]);
    }
    return 0;
}