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
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：
1 int merge(int u,int v) {
2 if (!u) return v;
3 if (!v) return u;
4 ch[u][0]=merge(ch[u][0],ch[v][0]);
5 ch[u][1]=merge(ch[u][1],ch[v][1]);
6 return u;
7 }
那么这题差不多就可以A了。
code：
1 %:pragma gcc optimize(3)
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<vector>
6 #define jug(i,x) (((1<<i)&x)>0)
7 #define M(a,x) memset(a,x,sizeof a)
8 using namespace std;
9 const int N=100005,Nod=12000005;
10 int n,tot,Q,a[N],lnk[N],nxt[N],son[N];
11 int ro[N],ans[N];
12 struct que {int v,i;};
13 vector <que> qr[N];
14 struct persistent_trie {
15 int cnt,ch[Nod][2];
16 void init() {cnt=1;}
17 int newnode() {
18 M(ch[cnt],0);
19 return cnt++;
20 }
21 int merge(int x,int y) {
22 if (!x) return y;
23 if (!y) return x;
24 ch[x][0]=merge(ch[x][0],ch[y][0]);
25 ch[x][1]=merge(ch[x][1],ch[y][1]);
26 return x;
27 }
28 void insert(int x,int v) {
29 int u=ro[x];
30 for (int i=30; i>=0; i--) {
31 bool c=jug(i,v);
32 if (!ch[u][c]) ch[u][c]=newnode();
33 u=ch[u][c];
34 }
35 }
36 int query(int x,int v) {
37 int u=ro[x],ret=0;
38 for (int i=30; i>=0; i--) {
39 bool c=jug(i,v);
40 if (ch[u][1-c]) ret|=(1<<i),u=ch[u][1-c];
41 else u=ch[u][c];
42 }
43 return ret;
44 }
45 }pt;
46 inline int read() {
47 int x=0; char ch=getchar();
48 while (ch<‘0‘||ch>‘9‘) ch=getchar();
49 while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
50 return x;
51 }
52 void add(int x,int y) {
53 nxt[++tot]=lnk[x],son[tot]=y,lnk[x]=tot;
54 }
55 void DFS(int x) {
56 ro[x]=pt.newnode();
57 pt.insert(x,a[x]);
58 for (int j=lnk[x]; j; j=nxt[j])
59 DFS(son[j]),ro[x]=pt.merge(ro[x],ro[son[j]]);
60 for (int i=0,si=qr[x].size(); i<si; i++)
61 ans[qr[x][i].i]=pt.query(x,qr[x][i].v);
62 }
63 int main() {
64 while (scanf("%d%d",&n,&Q)!=EOF) {
65 tot=0,M(lnk,0),M(nxt,0),M(ans,0),pt.init();
66 for (int i=1; i<=n; i++) a[i]=read();
67 for (int i=1; i<n; i++) {
68 int x=read(); add(x,i+1);
69 }
70 for (int i=1; i<=n; i++) qr[i].clear();
71 for (int i=1; i<=Q; i++) {
72 int x=read(); que now;
73 now.i=i,now.v=read(),qr[x].push_back(now);
74 }
75 DFS(1);
76 for (int i=1; i<=Q; i++) printf("%d\n",ans[i]);
77 }
78 return 0;
79 }