TeaTree

9022: TeaTree

时间限制: 3 Sec  内存限制: 1024 MB
提交: 15  解决: 8
[提交] [状态] [命题人:admin]

题目描述

Recently, TeaTree acquire new knoledge gcd (Greatest Common Divisor), now she want to test you.
As we know, TeaTree is a tree and her root is node 1, she have n nodes and n-1 edge, for each node i, it has it’s value v[i].
For every two nodes i and j (i is not equal to j), they will tell their Lowest Common Ancestors (LCA) a number : gcd(v[i],v[j]).
For each node, you have to calculate the max number that it heard. some definition:
In graph theory and computer science, the lowest common ancestor (LCA) of two nodes u and v in a tree is the lowest (deepest) node that has both u and v as descendants, where we define each node to be a descendant of itself.

输入

On the first line, there is a positive integer n, which describe the number of nodes.
Next line there are n-1 positive integers f[2] ,f[3], …, f[n], f[i] describe the father of node i on tree.
Next line there are n positive integers v[2] ,v[3], …, v[n], v[i] describe the value of node i.
n<=100000, f[i]<i, v[i]<=100000

输出

Your output should include n lines, for i-th line, output the max number that node i heard.
For the nodes who heard nothing, output -1.

样例输入

复制样例数据

4
1 1 3
4 1 6 9

样例输出

2
-1
3
-1

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e5 + 10000;
int n,root[maxn],ls[maxn*400],rs[maxn*400],tot,w[maxn*400],res[maxn];
vector<int>v[maxn],re[maxn];

inline void init(){
    for(register int i=1;i<maxn-108;++i){
        for(register int j=1;j<=sqrt(i);++j){
            if(i%j==0){
                v[i].emplace_back(j);
                if(j!=i/j){
                    v[i].emplace_back(i/j);
                }
            }
        }
    }
}
inline void update(int l,int r,int val,int &id){
    if(!id)id=++tot;
    if(l==r){
        w[id]=val;
        return;
    }
    int mid=l+r>>1;
    if(val<=mid)update(l,mid,val,ls[id]);
    else update(mid+1,r,val,rs[id]);
    w[id]=max(w[ls[id]],w[rs[id]]);
}
inline int merge(int fa,int v,int &ans){
    if(!fa||!v)return fa|v;
    if(w[fa]==w[v])ans=max(ans,w[fa]);
    if(ls[fa]||ls[v])ls[fa]=merge(ls[fa],ls[v],ans);
    if(rs[fa]||rs[v])rs[fa]=merge(rs[fa],rs[v],ans);
    return fa;
}
inline void dfs(int cur){
    for(register int i=0;i<re[cur].size();++i){
        int to=re[cur][i];
        dfs(to);
        merge(root[cur],root[to],res[cur]);
    }
}
int main() {
    //freopen("1.txt", "r", stdin);
    init();
    scanf("%d",&n);
    //printf("***");
    for(register int i=2,x;i<=n;++i){
        scanf("%d",&x);
        re[x].emplace_back(i);
    }
    for(register int i=1,val;i<=n;++i){
        scanf("%d",&val);
        for(register int j=0;j<v[val].size();++j){
            update(1,1e5,v[val][j],root[i]);
        }
    }
    for(register int i=1;i<=n;++i)res[i]=-1;
    dfs(1);
    for(register int i=1;i<=n;++i)printf("%d
",res[i]);
    return 0;
}