Weak Pair

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 5985    Accepted Submission(s): 1683

 

Problem Description

You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a non-negative value ai is assigned.An ordered pair of nodes (u,v) is said to be weak if
  (1) u is an ancestor of v (Note: In this problem a node u is not considered an ancestor of itself);
  (2) au×av≤k.

Can you find the number of weak pairs in the tree?

 

 

Input

There are multiple cases in the data set.
  The first line of input contains an integer T denoting number of test cases.
  For each case, the first line contains two space-separated integers, N and k, respectively.
  The second line contains N space-separated integers, denoting a1 to aN.
  Each of the subsequent lines contains two space-separated integers defining an edge connecting nodes u and v , where node u is the parent of node v.

  Constrains: 
  
  1≤N≤105 
  
  0≤ai≤109 
  
  0≤k≤1018

 

 

Output

For each test case, print a single integer on a single line denoting the number of weak pairs in the tree.

 

 

Sample Input


 


1 2 3 1 2 1 2

 

 

Sample Output


 


1

 

 

Source

​​2016 ACM/ICPC Asia Regional Dalian Online​​

 

 

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题意：

weak pair的要求：

1.u是v的祖先（注意不一定是父亲）

2.val[u]*val[v] <=k;

问有多对？

分析：

val[u]<=k/val[v],对于当前节点来说，即祖先节点有多少满足小于 k/val[u]，利用dfs的特性建立树状数组，一看代码就懂。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long long ll;
typedef  unsigned long long ULL;
const int N=500005;
LL val[N],c[N];
int in[N];
vector<int>G[N];
vector<LL>v;
int n,m;
LL k;
int lowbit(int x)
{
    return x & -x;
}
ll sum(int x)    //树状数组 前缀和查询
{
    ll sum = 0;
    while(x)
    {
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}
void add(int x, ll val) //树状数组 单点修改
{
    while(x <= m)
    {
        c[x] += val;
        x += lowbit(x);
    }
}

LL ans=0;
int getid(LL x)
{
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
void dfs(int u)
{

    for(int i=0; i<G[u].size(); i++)
    {
        int v=G[u][i];
        if(val[v]==0)
            ans+=sum(getid(1e18));
        else
            ans+=sum(getid(k/val[v]));

        add(getid(val[v]),1);
        dfs(v);
        add(getid(val[v]),-1);

    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(c,0,sizeof(c));
        v.clear();
        ans=0;

        scanf("%d%lld",&n,&k);


        for(int i=1; i<=n; i++)
        {
            in[i]=0;
            G[i].clear();
            scanf("%lld",&val[i]);
            v.push_back(val[i]);
            if(val[i]==0)
            {
                v.push_back(1e18);
            }
            else
                v.push_back(k/val[i]);
        }
        sort(v.begin(),v.end());
        v.erase(unique(v.begin(),v.end()),v.end());
        m=v.size();



        int u,v;
        for(int i=1; i<n; i++)
        {
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
            in[v]++;
        }

        int root=1;
        for(int i=1; i<=n; i++)
        {
            if(in[i]==0)
            {
                root=i;
                break;
            }
        }

        if(val[root]==0)
            add(getid(1e18),1);
        else
        {

            add(getid(val[root]),1);
        }

        //cout<<ans<<endl;
        dfs(root);
        printf("%lld\n",ans);

    }

    return 0;
}