HDU5877 Weak pair 树状数组+dfs

本题可以看出是算number的个数，重点是转化父子节点的关系，我们可以通过dfs来获得前面需要的节点

注意的是，这个点加完计算后，一定要删除，不然会影响兄弟节点的计算

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map> 
#include<set>
#include<vector> 
using namespace std;
typedef  long long ll;
const int N=5e5+10;
const ll inf=0x3f3f3f3f;
int h[N],ne[N],e[N],in[N];
int idx;
int a[N];
int b[N];
ll n;
int tr[N];
int size;
ll k;
ll ans;
vector<int> num;
void add(int a,int b){
    e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
int lowbit(int x){
    return x&-x;
}
void modify(int x,int c){
    int i;
    for(i=x;i<=n;i+=lowbit(i)){
        tr[i]+=c;
    }
}
ll sum(int x){
    int i;
    ll res=0;
    for(i=x;i;i-=lowbit(i)){
        res+=tr[i];
    }
    return res;
}
void dfs(ll u){
    int i;
    int x=upper_bound(num.begin(),num.end(),k/a[u])-num.begin();
    ans+=sum(x);
    int y=lower_bound(num.begin(),num.end(),a[u])-num.begin()+1;
    modify(y,1);
    for(i=h[u];i!=-1;i=ne[i]){
        int j=e[i];
        dfs(j);
    }
    modify(y,-1);
}
int main(){
    int t;
    cin>>t;
    while(t--){
        idx=0;
        ans=0;
        num.clear();
        scanf("%d%lld",&n,&k);
        memset(h,-1,sizeof h);
        memset(in,0,sizeof in);
        memset(tr,0,sizeof tr);
    int i;
    for(i=1;i<=n;i++){
     scanf("%d",&a[i]);    
      num.push_back(a[i]);
    }
    sort(num.begin(),num.end());
    num.erase(unique(num.begin(),num.end()),num.end());
    for(i=1;i<=n-1;i++){
       ll b,c;
       scanf("%lld%lld",&c,&b);
       add(c,b);
       in[b]++;    
    }
    for(i=1;i<=n;i++){
        if(!in[i]){
            dfs(i);
            break;
        }
    }
    cout<<ans<<endl;
    }
    return 0;
}