Problem E. TeaTree
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 28 Accepted Submission(s): 11
Problem Description
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.
Input
On 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]<=100000
Output
Your 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
Source
2018 Multi-University Training Contest 10
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chendu
题意：有一棵树，每对节点都会告诉他的最近公共祖先他们的GCD，求每一个节点知道的最大的GCD。
解题思路：先预处理，把每一个节点的所有因子预处理出来。然后对树深搜，回溯的时候记录这颗子树的所有因子，然后做启发式合并，小的并入大的。然后更新最大值即可。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAXN=200005;
const ll MOD=998244353;
vector<int> ch[100005];
int w[100005];
int ans[100005];
set<int> s[100005];
vector<int> yinzi[100005];
int merge(int x,int y){
int res=-1;
if(s[x].size()<s[y].size())
swap(s[x],s[y]);
for(set<int>::iterator it = s[y].begin();it!=s[y].end();it++){
if(s[x].count(*it)){
res=max(res,*it);
}
else{
s[x].insert(*it);
}
}
s[y].clear();
return res;
}
void dfs(int x){
for(int i=0;i<ch[x].size();i++){
int j=ch[x][i];
dfs(j);
ans[x]=max(ans[x],merge(x,j));
}
for(int i=0;i<yinzi[w[x]].size();i++){
if(s[x].count(yinzi[w[x]][i])){
ans[x]=max(ans[x],yinzi[w[x]][i]);
}
else{
s[x].insert(yinzi[w[x]][i]);
}
}
}
int main(){
//预处理因子
for(int i=1;i<=100000;i++)
for(int j=i;j<=100000;j+=i)
yinzi[j].push_back(i);
int N;
while(~scanf("%d",&N)){
memset(ans,-1,sizeof(ans));
int temp;
for(int i=2;i<=N;i++){
scanf("%d",&temp);
ch[temp].push_back(i);
}
for(int i=1;i<=N;i++)
scanf("%d",&w[i]);
dfs(1);
for(int i=1;i<=N;i++)
printf("%d\n",ans[i]);
for(int i=1;i<=N;i++){
ch[i].clear();
s[i].clear();
}
}
return 0;
}