hdu 5877 Weak Pair dfs序+树状数组+离散化

Weak Pair

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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

题意：给你一颗树，给点的权值，和树的边，求每个点v的祖先u，并且a[u]*a[v]<=K；

思路：利用dfs序可以快速的得到每个点的祖先，每次更新a[u]，找到k/a[v]>=a[u]的个数树状数组优化；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=1e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
vector<int>v[N];
int du[N];
ll l[N<<1],a[N];
int n,len;
ll k,ans;
int tree[N<<1];
void init(int n)
{
    for(int i=1;i<=n;i++)
        v[i].clear();
    memset(tree,0,sizeof(tree));
    memset(du,0,sizeof(du));
    ans=0;
}
int getpos(ll x)
{
    int pos=lower_bound(l,l+len,x)-l;
    return pos+1;
}
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int c)
{
    while(x<(N<<1))
    {
        tree[x]+=c;
        x+=lowbit(x);
    }
}
ll query(int x)
{
    ll ans=0;
    while(x)
    {
        ans+=tree[x];
        x-=lowbit(x);
    }
    return ans;
}
void dfs(int u)
{
    int p,q;
    if(a[u])
        p=getpos(k/a[u]);
    else
        p=(N<<1)-1;
    if(a[u])
        q=getpos(a[u]);
    else
        q=1;
    ans+=query(p);
    update(q,1);
    for(int i=0;i<v[u].size();i++)
    {
        dfs(v[u][i]);
    }
    update(q,-1);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int flag=0;
        scanf("%d%lld",&n,&k);
        init(n);
        for(int i=1;i<=n;i++)
        scanf("%lld",&a[i]),l[flag++]=k/a[i],l[flag++]=a[i];
        sort(l,l+flag);
        len=unique(l,l+flag)-l;
        for(int i=1;i<n;i++)
        {
            int u,w;
            scanf("%d%d",&u,&w);
            v[u].push_back(w);
            du[w]++;
        }
        for(int i=1;i<=n;i++)
            if(du[i]==0)
                dfs(i);
        printf("%lld\n",ans);
    }
    return 0;
}