2017ACM/ICPC广西邀请赛 K- Query on A Tree trie树合并
Query on A Tree
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
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?
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
1 2
1
1 3
2 1
题解:
每个数存在各自trie树里边,n个点这是棵树,再从底向上tri树合并起来
查询就是查询一颗合并后的trie树,利用从高位到低位,贪心取
- #include <bits/stdc++.h>
- inline int read(){int x=,f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}return x*f;}
- using namespace std;
- #define LL long long
- const int N = 2e5;
- vector<int > G[N];
- int n, q, x, u, a[N];
- int ch[N*][], root[N],sz;
- void inserts(int u,int x) {
- root[u] = ++sz;
- int tmp = sz;
- int y = sz;
- for(int i = ; i >= ; --i) {
- int tmps = (x>>i)&;
- if(!ch[y][tmps]) ch[y][tmps] = ++sz;
- y = ch[y][tmps];
- }
- }
- int merges(int u,int to) {
- if(u == ) return to;
- if(to == ) return u;
- int t = ++sz;
- ch[t][] = merges(ch[u][],ch[to][]);
- ch[t][] = merges(ch[u][],ch[to][]);
- return t;
- }
- void dfs(int u) {
- inserts(u,a[u]);
- for(auto to:G[u]) {
- dfs(to);
- root[u] = merges(root[u],root[to]);
- }
- }
- LL query(int u,int x) {
- int y = root[u];
- LL ret = ;
- for(int i = ; i >= ; --i) {
- int tmps = (x>>i)&;
- if(ch[y][tmps^]) ret += (<<i),y = ch[y][tmps^];
- else y = ch[y][tmps];
- }
- return ret;
- }
- void init() {
- for(int i = ; i <= n; ++i) root[i] = ,G[i].clear();
- sz = ;
- memset(ch,,sizeof(ch));
- }
- int main( int argc , char * argv[] ){
- while(scanf("%d%d",&n,&q)!=EOF) {
- for(int i = ; i <= n; ++i) scanf("%d",&a[i]);
- init();
- for(int i = ; i <= n; ++i) {
- scanf("%d",&x);
- G[x].push_back(i);
- }
- dfs();
- for(int i = ; i <= q; ++i) {
- scanf("%d%d",&u,&x);
- printf("%lld\n",query(u,x));
- }
- }
- return ;
- }
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