HDU-6430：Problem E. TeaTree（bitset）

Problem E. TeaTree
Time Limit: 8000/4000 MS (Java/Others)
Memory Limit: 524288/524288 K (Java/Others)

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] 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]) g c d ( 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] 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] v [ 2 ] , v [ 3 ] , … , v [ n ] , v [ i ] describe the value of node i.
n<=100000,f[i]<i,v[i]<=100000 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

思路：因为每个数的因子只有100多个，所以看似暴力的做法其实可以过。从大神那学习了一种写法，bitset动态存储每个子树的所有的因子数。

把每个数当成一个bit，然后把约数反向存入bitset，这样就可以用bitset的_Find_first()函数找到最低位的1，即最大的约数。

PS：如果这棵树是一条链的话，空间可能就炸了。所以还是建议用启发式合并。。

#include
using namespace std;
const int MAX=1e5+10;
const int N=1e5;
typedef long long ll;
vector<int>e[MAX];
vector<int>d[MAX];
int a[MAX],ans[MAX];
bitset*b[MAX];
void dfs(int k)
{
    b[k]=new bitset(0);//动态开内存
    for(int j=0;j1;//将a[k]的约数反着存入节点
    for(int i=0;iint nex=e[k][i];
        dfs(nex);
        int x=((*b[k])&(*b[nex]))._Find_first();//找到bitset里面最低位的1
        (*b[k])|=(*b[nex]); //合并儿子节点
        ans[k]=max(ans[k],N-x);
        delete b[nex];      //删除儿子节点
    }
}
int main()
{
    for(int i=1;i<=1e5;i++)
    for(int j=i;j<=1e5;j+=i)d[j].push_back(i);//预处理约数
    int n;
    scanf("%d",&n);
    for(int i=2;i<=n;i++)
    {
        int x;
        scanf("%d",&x);
        e[x].push_back(i);
    }
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    memset(ans,-1,sizeof ans);
    dfs(1);
    for(int i=1;i<=n;i++)printf("%d\n",ans[i]);
    return 0;
}