mission ini common enc deep edge tree def acc

Problem E. TeaTree

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 722 Accepted Submission(s): 255

Problem Description 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.

Input 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

Output 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. 求樹上每點的一個值 這個值 是 該點 以及它的子樹所有點的 最大gcd

#include <iostream>
#include <vector>

#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=a-1;i>=b;i--)

const int MX = 1e5;
const int MXX = 400*MX;
using namespace std;

int n;

vector<int> G[MX+5],vv[MX+5];


// init函數 實現將i因數分解 （i < 1e5）
void init() {
    rep(i,1,MX+1) vv[i].push_back(1);
    rep(i,2,MX+1) {
        vv[i].push_back(i);
        for(int j=i+i;j<=MX;j+=i) vv[j].push_back(i);
    } 

    // rep(i,1,MX+1) {
    //     rep(j,0,vv[i].size()) {
    //         printf("%d ",vv[i][j]);
    //     }puts("");
    // }
}

int root[MX+5],ls[MXX],rs[MXX],sum[MXX],rear,ans[MX];

inline void push_up(int rt) {
    if(ls[rt] && rs[rt]) sum[rt] = max(sum[ls[rt]],sum[rs[rt]]);
    else if(ls[rt]) sum[rt] = sum[ls[rt]];
    else if(rs[rt]) sum[rt] = sum[rs[rt]];
}

void update(int &rt,int l,int r,int p) {
    if(rt==0) rt = ++rear;
    if(l == r) {
        sum[rt] = p;
        return ;
    }
    int m = (l+r)>>1;
    if(p <= m) update(ls[rt],l,m,p);
    else update(rs[rt],m+1,r,p);
    push_up(rt);
}

int merge(int rt, int prt, int &ans) {
        if(rt==0 || prt==0) return rt^prt;
        //這裏維護最大的gcd
        if(sum[rt] == sum[prt]) ans = max(ans,sum[rt]);
        //這裏只有有因子，就歸並到rt上面
        if(ls[rt] | ls[prt]) ls[rt] = merge(ls[rt], ls[prt], ans);
        if(rs[rt] | rs[prt]) rs[rt] = merge(rs[rt], rs[prt], ans);
        push_up(rt);
        return rt;
}

void dfs(int u) {
    ans[u] = -1;
    rep(i, 0, G[u].size()) {
        int v = G[u][i];
        dfs(v);
        root[u] = merge(root[u],root[v],ans[u]);
    }
}

int main () {
    freopen("in.txt" ,"r",stdin);
    freopen("out.txt","w",stdout);
    init();
    
    scanf("%d", &n);
    //建邊
    rep(i,2,n+1) {
        int fa; scanf("%d",&fa);
        G[fa].push_back(i);
    }
    //對每個v[i]建線段樹
    rear=0;
    rep(i,1,n+1) {
        int x; scanf("%d", &x);
        root[i]=0;
        rep(j, 0, vv[x].size()) {
            update(root[i], 1, MX, vv[x][j]);
        }
    }
    //暴力更新 gcd
    dfs(1);
    //輸出答案
    rep(i,1,n+1) printf("%d\n", ans[i]);
    return 0;
}

hdu 6430 線段樹 暴力維護