CF 题目集锦 PART 7 #264 div 2 E

【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

4 610 8 4 31 22 33 41 11 21 31 42 1 91 4

output

-112-11

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_backusing 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 arr{int 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;}