线段树合并

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.

InputOn 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]<=100000OutputYour 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.Sample Input

4
1 1 3
4 1 6 9

Sample Output

2
-1
3
-1

题意 ： 给你一颗树，求以任意一个点为根节点时，该子树内最大的 gcd 是多少，并且所求gcd的两个点的 lca 等于根节点
思路分析 ： 1 . 用 vector 模拟归并排序，预处理一下每个数字的约数有哪些，然后合并就可以了

#define ll long long
const int maxn = 1e5+5;

int n;
vector<int>d[maxn], ve[maxn], f[maxn];
void init(){
    for(int i = 1; i <= 100000; i++){
        for(int j = i; j <= 100000; j += i){
            d[j].push_back(i);
        }
    } 
}
int val[maxn], ans[maxn];
vector<int>temp;

void merge(int x, int y){
    temp.clear();    
    if (f[x].size() == 0) f[x] = d[val[x]];
    if (f[y].size() == 0) f[y] = d[val[y]];
    
    int i = 0, j = 0;
    while((i < f[x].size()) && (j < f[y].size())){
        if (f[x][i] < f[y][j]) temp.push_back(f[x][i]), i++;
        else if (f[x][i] > f[y][j]) temp.push_back(f[y][j]), j++;
        else {
            temp.push_back(f[x][i]);
            ans[x] = max(ans[x], f[x][i]); i++, j++;
        }
    }
    while (i < f[x].size()) {
        temp.push_back(f[x][i]);
        i++;
    }
    while(j < f[y].size()) {
        temp.push_back(f[y][j]);
        j++;
    }
    f[x].clear(); f[y].clear();
    f[x] = temp; 
}


void merge8(int x,int v)
{
    int i = 0, j = 0;
    vector<int>tem;
    for(; i < f[x].size() && j < f[v].size();)
    {
        int a1 = f[x][i], a2 = f[v][j];
        if(a1 < a2)
        {
            tem.push_back(a1);
            i++;
        }
        else if(a1 == a2)
        {
            ans[x] = max(ans[x],a1);
            tem.push_back(a1); 
            i++;j++; 
        }
        else
        {
            tem.push_back(a2); 
            j++;
        }
    }
    for(; i < f[x].size(); i++)tem.push_back(f[x][i]);
    for(; j < f[v].size(); j++)tem.push_back(f[v][j]);
    f[x] = tem;
    return;
}

void dfs(int x){
    for(int i = 0; i < ve[x].size(); i++){
        int to = ve[x][i];
        dfs(to);
        merge(x, to);
    }
}

int main() {
    init();
    cin >> n;
    
    int x;
    memset(ans, -1, sizeof(ans));
    for(int i = 2; i <= n; i++){
        scanf("%d", &x);
        ve[x].push_back(i);
    }
    for(int i = 1; i <= n; i++){
        scanf("%d", &val[i]);
        //f[i] = d[val[i]]; 
    }
    dfs(1);    
    for(int i = 1; i <= n; i++) printf("%d
", ans[i]);
    return 0;
}

 2 . 对每个权值建立一颗权值线段树，10万以内的数不同的约数最多有 400  个，然后将线段树两两合并即可

代码示例 ：

#define ll long long
const int maxn = 1e5+5;

int n;
vector<int>ve[maxn], d[maxn];
int root[maxn], cnt = 1;
int lson[maxn*400], rson[maxn*400], maxx[maxn*400];

void init(){
    for(int i = 1; i <= 100000; i++){
        for(int j = i; j <= 100000; j += i) 
            d[j].push_back(i);
    } 
}

void pushup(int k){
    if (lson[k] && rson[k]) maxx[k] = max(maxx[lson[k]], maxx[rson[k]]);
    else if (lson[k]) maxx[k] = maxx[lson[k]];
    else if (rson[k]) maxx[k] = maxx[rson[k]];
}

void update(int &rt, int l, int r, int num){
    if (!rt) rt = cnt++;
    if (l == r) { maxx[rt] = num; return; }
    int mid = (l+r)>>1;
    
    if (num <= mid) update(lson[rt], l, mid, num);
    else update(rson[rt], mid+1, r, num);
    pushup(rt);
}

int merge(int x, int y, int &ans){ 
    if (!x || !y) return x^y;
    
    if (maxx[x] == maxx[y]) ans = max(ans, maxx[x]);
    if (lson[x] || lson[y]) lson[x] = merge(lson[x], lson[y], ans);
    if (rson[x] || rson[y]) rson[x] = merge(rson[x], rson[y], ans);
    
    pushup(x);
    return x;
}

int ans[maxn];
void dfs(int x){
    ans[x] = -1;
    
    for(int i = 0; i < ve[x].size(); i++){
        int to = ve[x][i];
        dfs(to);
        root[x] = merge(root[x], root[to], ans[x]);
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int x;
    init();
    
    cin >> n;
    for(int i = 2; i <= n; i++){
        scanf("%d", &x);
        ve[x].push_back(i);
    } 
    int val;
    for(int i = 1; i <= n; i++){
        root[i] = 0;
        scanf("%d", &val);
        for(int j = 0; j < d[val].size(); j++){
            update(root[i], 1, 100000, d[val][j]);
        }        
    }
    dfs(1);
    for(int i = 1; i <= n; i++) printf("%d
", ans[i]);
    return 0;
}