1 //树形DP+树状数组 HDU 5877  Weak Pair 2 // 思路：用树状数组每次加k/a[i]，每个节点ans+=Sum(a[i]) 表示每次加大于等于a[i]的值 3 // 这道题要离散化 4  5 #include <bits/stdc++.h> 6 using namespace std; 7 #define LL long long 8 typedef pair<int,int> pii; 9 const double inf = 123456789012345.0;10 const LL MOD =100000000LL;11 const int N = 2e5+10;12 const int maxx = 200010; 13 #define clc(a,b) memset(a,b,sizeof(a))14 const double eps = 1e-7;15 void fre() {freopen("in.txt","r",stdin);}16 void freout() {freopen("out.txt","w",stdout);}17 inline int read() {int x=0,f=1;char ch=getchar();while(ch>‘9‘||ch<‘0‘) {if(ch==‘-‘) f=-1; ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘) {x=x*10+ch-‘0‘;ch=getchar();}return x*f;}18 19 map<LL,LL> ma;20 LL a[N];21 LL c[N],b[N];22 LL in[N];23 vector<LL> g[N];24 LL lowbit(LL x){ return x&(-x);}25 LL add(LL x,int t){26     while(x>0){27        c[x]+=t;28        x-=lowbit(x);29     }30 }31 LL Sum(LL x){32     LL sum=0;33     while(x<maxx){34         sum+=c[x];35         x+=lowbit(x);36     }37     return sum;38 }39 40 LL ans=0;41 LL n,k;42 void dfs(LL rt){43      for(LL i=0;i<(int)g[rt].size();i++){44          LL v=g[rt][i];45          ans+=Sum(ma[a[v]]);46          if(a[v]==0) add(maxx,1);47          else add(ma[k/a[v]],1);48          dfs(v);49          if(a[v]==0) add(maxx,-1);50          else add(ma[k/a[v]],-1);51      }52 }53 int main(){54     int T;55     scanf("%d",&T);56     while(T--){57         ma.clear();58         memset(c,0,sizeof(c));59         scanf("%I64d%I64d",&n,&k);60         for(int i=1;i<=n;i++){61             scanf("%I64d",&a[i]);62             b[i*2-2]=a[i];63             if(a[i]!=0) b[i*2-1]=k/a[i];64             g[i].clear();65             in[i]=0;66         }67         sort(b,b+2*n);68         int K=unique(b,b+2*n)-b;69         int cxt=0;70         for(int i=0;i<K;i++){71             ma[b[i]]=++cxt;72         }73         for(LL i=0;i<n-1;i++){74             LL u,v;75             scanf("%I64d%I64d",&u,&v);76             g[u].push_back(v);77             in[v]++;78         }79         LL rt;80         for(LL i=1;i<=n;i++){81             if(in[i]==0){82                 rt=i;83                 break;84             }85         }86         ans=0;87         if(a[rt]==0) add(maxx,1);88         else add(ma[k/a[rt]],1);89         dfs(rt);90         printf("%I64d\n",ans);91     }92     return 0;93 }