杭电多校第六场-J-Ridiculous Netizens

Problem Description

Mr. Bread has a tree T with n vertices, labeled by 1,2,…,n. Each vertex of the tree has a positive integer value wi.

The value of a non-empty tree T is equal to w1×w2×⋯×wn. A subtree of T is a connected subgraph of T that is also a tree.

Please write a program to calculate the number of non-empty subtrees of T whose values are not larger than a given number m.

Input

The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.

In each test case, there are two integers n,m(1≤n≤2000,1≤m≤106) in the first line, denoting the number of vertices and the upper bound.

In the second line, there are n integers w1,w2,…,wn(1≤wi≤m), denoting the value of each vertex.

Each of the following n−1 lines contains two integers ui,vi(1≤ui,vi≤n,ui≠vi), denoting an bidirectional edge between vertices ui and vi.

Output

For each test case, print a single line containing an integer, denoting the number of valid non-empty subtrees. As the answer can be very large, output it modulo 10^9+7.

Sample Input

1

5 6

1 2 1 2 3

1 2

1 3

2 4

2 5

Sample Output

14

 

题意：
一棵无根树，每个点有权值，询问有多少个联通子图的权值的积等于m

思路
https://www.cnblogs.com/hua-dong/p/11320013.html
考虑对某点，联通块要么经过它要么不经过它 ——> 点分治
对于经过该点的用dp求解 
在dfs序上dp，类似于树形依赖背包
dp[i][j]表示 dfs序i之后的乘积为j的方案数
可知 dp[i][j]=(dp[i+1][j/a[dfn[i]]]+dp[i+son[i]][j]) //当前点选/不选
但第二维为m不可行
考虑把<sqrt(M)的和大于sqrt(M)的分开保存，那么前者就是正常的背包，表示当前乘积；后者可以看成以后还可以取多少

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
const int N=1e5+10;
const int p=1e9+7;
int T,n,m,cnt,sum,root,tim,ans;
struct orz{
    int v,nex;}e[N*2];
int a[N],last[N],son[N],f[N],dfn[N],dp1[N][1005],dp2[N][1005];
bool vis[N];
void add(int x,int y)
{
    cnt++;
    e[cnt].v=y;
    e[cnt].nex=last[x];
    last[x]=cnt;
}
void getroot(int x,int fa)
{
    son[x]=1; f[x]=0;
    for (int i=last[x];i;i=e[i].nex)
    {
        if (e[i].v==fa || vis[e[i].v]) continue;
        getroot(e[i].v,x);
        son[x]+=son[e[i].v];
        f[x]=max(f[x],son[e[i].v]);
    }
    f[x]=max(f[x],sum-son[x]);
    if (f[x]<f[root]) root=x;
}
void dfs(int x,int fa)
{
    dfn[++tim]=x; son[x]=1;
    for (int i=last[x];i;i=e[i].nex)
    {
        if (e[i].v==fa || vis[e[i].v]) continue;
        dfs(e[i].v,x);
        son[x]+=son[e[i].v];
    }
}
void cal()
{
    int mm=sqrt(m);
    for (int i=1;i<=tim+1;i++)
    {
        memset(dp1[i],0,sizeof(dp1[i]));
        memset(dp2[i],0,sizeof(dp2[i]));
    }
    dp1[tim+1][1]=1;
    for (int i=tim;i>=1;i--)
    {
        int x=a[dfn[i]];
        for (int j=1;j<=min(mm,m/x);j++)
        {
            int k=j*x;
            if (k<=mm) dp1[i][k]=(dp1[i][k]+dp1[i+1][j])%p;
            else dp2[i][m/k]=(dp2[i][m/k]+dp1[i+1][j])%p;
        }
        for (int j=x;j<=mm;j++)
        {
            dp2[i][j/x]=(dp2[i][j/x]+dp2[i+1][j])%p;
        }
        for (int j=1;j<=mm;j++)
        {
            dp1[i][j]=(dp1[i][j]+dp1[i+son[dfn[i]]][j])%p;
            dp2[i][j]=(dp2[i][j]+dp2[i+son[dfn[i]]][j])%p;
        }
    }
    for (int i=1;i<=mm;i++)
    {
        ans=(ans+dp1[1][i])%p;
        ans=(ans+dp2[1][i])%p;
    }
    ans=(ans-1+p)%p;
}
void work(int x)
{
    //cout<<x<<endl;
    vis[x]=1; tim=0;
    dfs(root,0); //for (int i=1;i<=tim;i++) cout<<dfn[i]<<' '; cout<<endl;
    cal();
    for (int i=last[x];i;i=e[i].nex)
    {
        if (vis[e[i].v]) continue;
        sum=son[e[i].v];
        root=0;
        getroot(e[i].v,root);
        work(root);
    }
}
void init()
{
    cnt=0; ans=0;
    for (int i=1;i<=n;i++) last[i]=0,vis[i]=0;
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        init();
        for (int i=1;i<=n;i++) scanf("%d",&a[i]);
        int x,y;
        for (int i=1;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            add(x,y); add(y,x);
        }
        sum=n; f[0]=inf;
        getroot(1,0);
        work(root);
        printf("%d\n",ans);
    }
    return 0;
}