【原题】
10 seconds
256 megabytes
standard input
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.
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.
For each query of the first type output the result of the query.
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
-1 1 2 -1 1
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 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;
}