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

?

題意： 一棵無根樹，每個(gè)點(diǎn)有權(quán)值，詢問有多少個(gè)聯(lián)通子圖的權(quán)值的積等于m

思路
https://www.cnblogs.com/hua-dong/p/11320013.html
考慮對(duì)某點(diǎn)，聯(lián)通塊要么經(jīng)過它要么不經(jīng)過它 ——> 點(diǎn)分治
對(duì)于經(jīng)過該點(diǎn)的用dp求解
在dfs序上dp，類似于樹形依賴背包
dp[i][j]表示 dfs序i之后的乘積為j的方案數(shù)
可知 dp[i][j]=(dp[i+1][j/a[dfn[i]]]+dp[i+son[i]][j]) //當(dāng)前點(diǎn)選/不選
但第二維為m不可行
考慮把<sqrt(M)的和大于sqrt(M)的分開保存，那么前者就是正常的背包，表示當(dāng)前乘積；后者可以看成以后還可以取多少 #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; } View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/tetew/p/11366398.html

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