CF 题目集锦 PART 7 #264 div 2 E

Posted 2022-12-01 阿蒋

【原题】

E. Caisa and Tree time limit per test 10 seconds memory limit per test 256 megabytes input standard input output standard output

Caisa is now at home and his son has a simple task for him.

Given a rooted tree with n vertices, numbered from 1 to n (vertex 1 is the root). Each vertex of the tree has a value. You should answer q queries. Each query is one of the following:

• Format of the query is "1 v". Let's write out the sequence of vertices along the path from the root to vertex v: u1, u2, ..., uk (u1 = 1; uk = v). You need to output such a vertex ui that gcd(value of ui, value of v) > 1 and i < k. If there are several possible vertices ui pick the one with maximum value of i. If there is no such vertex output -1.
• Format of the query is "2 v w". You must change the value of vertex v to w.

You are given all the queries, help Caisa to solve the problem.

Input

The first line contains two space-separated integers n, q (1 ≤ n, q ≤ 105).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 2·106), where ai represent the value of node i.

Each of the next n - 1 lines contains two integers xi and yi (1 ≤ xi, yi ≤ n; xi ≠ yi), denoting the edge of the tree between vertices xi and yi.

Each of the next q lines contains a query in the format that is given above. For each query the following inequalities hold: 1 ≤ v ≤ n and 1 ≤ w ≤ 2·106. Note that: there are no more than 50 queries that changes the value of a vertex.

Output

For each query of the first type output the result of the query.

Sample test(s) input

4 6
10 8 4 3
1 2
2 3
3 4
1 1
1 2
1 3
1 4
2 1 9
1 4

output

-1
1
2
-1
1

Note

gcd(x, y) is greatest common divisor of two integers x and y.

【分析】这道题是做现场赛的。本来能A的，但是太紧张了=而且也不会用vector，边表搞的麻烦死了。

开始看到修改操作才50次、时间又松，真是爽！估计每次可以暴力重构这颗树，然后对于每个质因子记录最优值。

首先每次不能sqrt的效率枚举一个数的因子，我们可以预处理出每个数的所有质因子。（其实有更省空间的）

剩下来要解决的问题是：因为我是用dfs的，怎么把某个子树的信息在搜完后再去掉？（以免影响其他子树）HHD表示用vector一点也不虚。其实应该也可以用边表类似的思路，但是麻烦= =

【代码】

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#define N 100005
#define S 2000005
#define push push_back
#define pop pop_back
using namespace std;
vector<int>fac[S],f[S];
int data[N],ans[N],end[N],pf[S],deep[N];
int C,cnt,n,Q,i,x,y,opt;
struct arrint go,next;a[N*2];
inline void add(int u,int v)a[++cnt].go=v;a[cnt].next=end[u];end[u]=cnt;
inline void init()

  int H=2000000;
  for (int i=2;i<=H;i++)
    if (!pf[i])
    
      for (int j=i;j<=H;j+=i)
        fac[j].push(i),pf[j]=1;
    

void dfs(int k,int fa)

  int P=data[k];
  for (int i=0;i<fac[P].size();i++)
  
    int go=fac[P][i],temp=f[go].size();
    if (temp&&deep[f[go][temp-1]]>deep[ans[k]]) ans[k]=f[go][temp-1];
    f[go].push(k);
  
  for (int i=end[k];i;i=a[i].next)
    if (a[i].go!=fa)
      dfs(a[i].go,k);
  for (int i=0;i<fac[P].size();i++)
    f[fac[P][i]].pop();

inline void get_deep(int k,int fa)

  for (int i=end[k];i;i=a[i].next)
    if (a[i].go!=fa) deep[a[i].go]=deep[k]+1,get_deep(a[i].go,k);

int main()

  scanf("%d%d",&n,&Q);
  for (i=1;i<=n;i++)
    scanf("%d",&data[i]);
  for (i=1;i<n;i++)
    scanf("%d%d",&x,&y),add(x,y),add(y,x);
  init();deep[0]=-1;get_deep(1,0);
  memset(ans,0,sizeof(ans));dfs(1,0);
  while (Q--)
  
    scanf("%d%d",&opt,&x);
    if (opt==1) printf("%d\\n",ans[x]?ans[x]:-1);continue;
    memset(ans,0,sizeof(ans));
    scanf("%d",&y);data[x]=y;dfs(1,0);
  
  return 0;