hdu 6191 Query on A Tree(字典树启发式合并（动态建树） 可持久化字典树+dfs序)

Query on A Tree

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 298    Accepted Submission(s): 116

Problem Description

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 21 211 32 1

 

Sample Output

23

这题主要是不会怎么将字典树给合并起来，其实只要将相同的值的子节点的地址赋给父节点地址就好了。。。

#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<vector>#include<map>#include <set>#include <bits/stdc++.h>using namespace std;const int N = 1e5+10;typedef long long LL;struct node{    int x,id;};typedef struct trie{    trie* next[2];}trie;int a[N], ans[N];vector<node> q[N];vector<int>p[N];trie *root[N];void update(trie *x,int vx){    trie *px=x;    for(int i=29;i>=0;i--)    {        int v=(vx&(1<<i))?1:0;        if(px->next[v]==NULL)        {            trie *tmp=new trie;            tmp->next[0]=tmp->next[1]=NULL;            px->next[v]=tmp;        }        px=px->next[v];    }    return ;}trie* merge1(trie *px,trie *qx){    if(px==NULL) return qx;    if(qx==NULL) return px;    px->next[0]=merge1(px->next[0],qx->next[0]);    px->next[1]=merge1(px->next[1],qx->next[1]);    free(qx);//释放q不然会超内存    return px;}int get(trie *x,int vx){    trie *px=x;    int r=0;    for(int i=29;i>=0;i--)    {        int v=(vx&(1<<i))?1:0;        if(px->next[1^v]!=NULL)        {            r|=(1<<i);            px=px->next[1^v];        }        else    px=px->next[v];    }    return r;}void dfs(int u){    root[u]=new trie;    root[u]->next[0]=root[u]->next[1]=NULL;    update(root[u],a[u]);    for(int i=0;i<p[u].size();i++)    {        int v=p[u][i];        dfs(v);        root[u]=merge1(root[u],root[v]);    }    for(int i=0;i<q[u].size();i++)    {        node tmp=q[u][i];        int h=get(root[u], tmp.x);        ans[tmp.id]=h;    }    return ;}void delete1(trie *x){    if(x->next[0]!=NULL) delete1(x->next[0]);    if(x->next[1]!=NULL) delete1(x->next[1]);    free(x);    return ;}int main(){    int n, m, x;    while(scanf("%d %d", &n, &m)!=EOF)    {        for(int i=1;i<=n;i++) scanf("%d", &a[i]),p[i].clear(),q[i].clear();        for(int i=2;i<=n;i++)        {            scanf("%d", &x);            p[x].push_back(i);        }        for(int i=1;i<=m;i++)        {            int x, y;            scanf("%d %d", &x, &y);            node tmp;            tmp.id=i,tmp.x=y;            q[x].push_back(tmp);        }        dfs(1);        for(int i=1;i<=m;i++) printf("%d\n",ans[i]);        delete1(root[1]);//释放内存    }    return 0;}

因为没有办法直接用字典树，所以给每一个节点进和出一个标号 然后就可以套模板了（中间变量名写错了，wa一上午，如果代码能力不稳的话，比赛时还是有一个人看着稳一点）

#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<vector>#include<map>#include <set>#include <bits/stdc++.h>using namespace std;const int N = 260005;typedef long long LL;int head[N];int a[N];struct node{    int to,next;}p[N*2];int cnt, num, tot;LL bit[40];int in[N], out[N], son[N*32][2], sum[N*32], rt[N];void init(){    memset(head,-1,sizeof(head));    memset(son,0,sizeof(son));    memset(sum,0,sizeof(sum));    cnt=0,num=0,tot=0;    return ;}void add(int u,int v){    p[cnt].to=v,p[cnt].next=head[u];    head[u]=cnt++;    return ;}void insert1(int &o,int pre,int x,int len){    o=++tot;    son[o][0]=son[pre][0], son[o][1]=son[o][1];    sum[o]=sum[pre]+1;    if(len==-1) return ;    int v=(x&(bit[len]))?1:0;    insert1(son[o][v],son[pre][v],x,len-1);}void dfs(int u,int fa){    in[u]=++num;    insert1(rt[in[u]],rt[in[u]-1],a[u],30);    for(int i=head[u];i!=-1;i=p[i].next)    {        int v=p[i].to;        if(v==fa) continue;        dfs(v,u);    }    out[u]=num;}int query(int ss,int tt,int x,int len){    if(len==-1) return 0;    int v=((x>>len)&1)?1:0;    if(sum[son[tt][1^v]]>sum[son[ss][1^v]])    {        return bit[len]+query(son[ss][1^v],son[tt][1^v],x,len-1);    }    return query(son[ss][v],son[tt][v],x,len-1);}int main(){    bit[0]=1;    for(int i=1;i<=30;i++) bit[i]=bit[i-1]*2;    int n, m;    while(scanf("%d %d", &n, &m)!=EOF)    {        init();        for(int i=1;i<=n;i++) scanf("%d", &a[i]);        for(int i=2;i<=n;i++)        {            int x;            scanf("%d", &x);            add(x,i);        }        dfs(1,0);        for(int i=1;i<=m;i++)        {            int u, x;            scanf("%d %d", &u, &x);            printf("%d\n",query(rt[in[u]-1],rt[out[u]],x,30));        }    }    return 0;}