权值线段树合并

线段树维护1～1e5这个值域，对于每个点开一颗线段树，储存值域内最大的因数。

然后对整个树dfs，合并父亲和儿子节点的线段树，在合并过程中更新答案。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int ret = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();    return w ? -ret : ret;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template 
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template 
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template 
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 100005;int n, cnt, tot, head[N], w[N], tree[(N*400)<<2], ls[(N*400)<<2], rs[(N*400)<<2], root[N], res[N];struct Edge{ int v, next; }edge[N<<1];vector
        
          fac[N];void calc(int x){    if(!fac[x].empty()) return;    fac[x].push_back(1), fac[x].push_back(x);    for(int i = 2; i <= sqrt(x) + 0.5; i ++){        if(x % i == 0){            fac[x].push_back(i), fac[x].push_back(x / i);        }    }}int build(){   tot ++;   tree[tot] = ls[tot] = rs[tot] = 0;   return tot;}void addEdge(int a, int b){    edge[cnt].v = b, edge[cnt].next = head[a], head[a] = cnt ++;}void push_up(int rt){    tree[rt] = max(tree[ls[rt]], tree[rs[rt]]);}void insert(int rt, int l, int r, int val){    if(l == r){        tree[rt] = l;        return;    }    int mid = (l + r) >> 1;    if(val <= mid){        if(!ls[rt]) ls[rt] = build();        insert(ls[rt], l, mid, val);    }    if(val > mid){        if(!rs[rt]) rs[rt] = build();        insert(rs[rt], mid + 1, r, val);    }    push_up(rt);}int merge(int p, int q, int l, int r, int &ans){    if(!p || !q) return p ^ q;    if(tree[p] == tree[q]) ans = max(ans, tree[p]);    if(l == r){        tree[p] = max(tree[p], tree[q]);        return p;    }    int mid = (l + r) >> 1;    ls[p] = merge(ls[p], ls[q], l, mid, ans);    rs[p] = merge(rs[p], rs[q], mid + 1, r, ans);    push_up(p);    return p;}void dfs(int s, int fa){    res[s] = -1;    for(int i = head[s]; i != -1; i = edge[i].next){        int u = edge[i].v;        if(u == fa) continue;        dfs(u, s);        root[s] = merge(root[s], root[u], 1, N, res[s]);    }}int main(){        full(head, -1);    n = read();    for(int i = 2; i <= n; i ++){        int v = read();        addEdge(v, i), addEdge(i, v);    }    for(int i = 1; i <= n; i ++) w[i] = read();    for(int i = 1; i <= n; i ++) calc(w[i]);    for(int i = 1; i <= n; i ++){        root[i] = build();        for(int j = 0; j < fac[w[i]].size(); j ++){            insert(root[i], 1, N, fac[w[i]][j]);        }    }    dfs(1, 0);    for(int i = 1; i <= n; i ++){        printf("%d\n", res[i]);    }    return 0;}