codeforces1101D GCD Counting 【树形DP】

题目分析：

蛮简单的一道题，对于每个数拆质因子，对于每个质因子找出最长链，在每个地方枚举一下拼接

代码：

#include<bits/stdc++.h>
using namespace std;

const int maxn = 205000;

int n,a[maxn],prime[maxn],flag[maxn],minn[maxn],num,ans;
vector <int> g[maxn];
vector <pair<int,int> > mp[maxn];

vector<int> cl[maxn];

void getprime(int N){
    for(int i=2;i<=N;i++){
    if(!flag[i]){prime[++num] = i,minn[i] = i;}
    for(int j=1;j<=num&&i*prime[j]<=N;j++){
        flag[i*prime[j]] = 1;
        minn[i*prime[j]] = prime[j];
        if(i%prime[j] == 0) break;
    }
    }
}

void read(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<n;i++){
    int u,v; scanf("%d%d",&u,&v);
    g[u].push_back(v); g[v].push_back(u);
    }
}

void dp(int now,int fa){
    for(int i=0;i<g[now].size();i++){
    if(g[now][i] == fa) continue;
    dp(g[now][i],now);
    }
    int p = a[now];
    while(p != 1){cl[minn[p]].clear(); p /= minn[p]; }
    for(int i=0;i<g[now].size();i++){
    if(g[now][i] == fa) continue;
    for(int j=0;j<mp[g[now][i]].size();j++){
        if(a[now] % mp[g[now][i]][j].first == 0){
        cl[mp[g[now][i]][j].first].push_back(mp[g[now][i]][j].second);
        }
    }
    }
    p = a[now];
    while(p != 1){
    int z = minn[p]; while(p%z == 0) p /= z;
    int maxx = 0,sec = 0;
    for(int i=0;i<cl[z].size();i++){
        if(cl[z][i] >= maxx) sec = maxx,maxx = cl[z][i];
        else if(cl[z][i] > sec) sec = cl[z][i];
    }
    mp[now].push_back(make_pair(z,maxx+1));
    ans = max(ans,maxx+sec+1);
    }
}

void work(){
    getprime(200000);
    dp(1,0);
    printf("%d
",ans);
}

int main(){
    read();
    work();
    return 0;
}