2018 Multi-University Training Contest 10

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 <bits/stdc++.h>
#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 <typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template <typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template <typename A, typename B, typename C>
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<int> 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
", res[i]);
    }
    return 0;
}