[hdu 6191] Query on A Tree

Query on A Tree

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 733    Accepted Submission(s): 275

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 2

1 2

1

1 3

2 1

 

Sample Output

2
3

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

显然，又是data structure。。。

网上有用DFS序来做的，但是本蒟蒻并不太清楚他们dalao的做法，所以只是大了一发可持久化trie合并。。

对于这一题，主要涉及trie的合并，要将u的子节点的信息合并到u的身上去。

那么，假设要将v的信息并到u上，则：

 int merge(int u,int v) {
     if (!u) return v;
     if (!v) return u;
     ch[u][]=merge(ch[u][],ch[v][]);
     ch[u][]=merge(ch[u][],ch[v][]);
     return u;
 }

那么这题差不多就可以A了。

code：

 %:pragma gcc optimize()
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #define jug(i,x) (((1<<i)&x)>0)
 #define M(a,x) memset(a,x,sizeof a)
 using namespace std;
 ,Nod=;
 int n,tot,Q,a[N],lnk[N],nxt[N],son[N];
 int ro[N],ans[N];
 struct que {int v,i;};
 vector <que> qr[N];
 struct persistent_trie {
     ];
     ;}
     int newnode() {
         M(ch[cnt],);
         return cnt++;
     }
     int merge(int x,int y) {
         if (!x) return y;
         if (!y) return x;
         ch[x][]=merge(ch[x][],ch[y][]);
         ch[x][]=merge(ch[x][],ch[y][]);
         return x;
     }
     void insert(int x,int v) {
         int u=ro[x];
         ; i>=; i--) {
             bool c=jug(i,v);
             if (!ch[u][c]) ch[u][c]=newnode();
             u=ch[u][c];
         }
     }
     int query(int x,int v) {
         ;
         ; i>=; i--) {
             bool c=jug(i,v);
             -c]) ret|=(<<i),u=ch[u][-c];
             else u=ch[u][c];
         }
         return ret;
     }
 }pt;
 inline int read() {
     ; char ch=getchar();
     ') ch=getchar();
     +ch-',ch=getchar();
     return x;
 }
 void add(int x,int y) {
     nxt[++tot]=lnk[x],son[tot]=y,lnk[x]=tot;
 }
 void DFS(int x) {
     ro[x]=pt.newnode();
     pt.insert(x,a[x]);
     for (int j=lnk[x]; j; j=nxt[j])
         DFS(son[j]),ro[x]=pt.merge(ro[x],ro[son[j]]);
     ,si=qr[x].size(); i<si; i++)
         ans[qr[x][i].i]=pt.query(x,qr[x][i].v);
 }
 int main() {
     while (scanf("%d%d",&n,&Q)!=EOF) {
         tot=,M(lnk,),M(nxt,),M(ans,),pt.init();
         ; i<=n; i++) a[i]=read();
         ; i<n; i++) {
             );
         }
         ; i<=n; i++) qr[i].clear();
         ; i<=Q; i++) {
             int x=read(); que now;
             now.i=i,now.v=read(),qr[x].push_back(now);
         }
         DFS();
         ; i<=Q; i++) printf("%d\n",ans[i]);
     }
     ;
 }