2016大连网络赛 Weak Pair

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

分析：dfs+树状数组+离散化；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=2e5+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,k,t,h[maxn],tot,q[maxn],fa[maxn],num;
ll ans,a[maxn],b[maxn],c[maxn];
struct node
{
    int to,nxt;
}e[maxn];
void add(int x,int y)
{
    tot++;
    e[tot].to=y;
    e[tot].nxt=h[x];
    h[x]=tot;
}
void gao(int x,int y)
{
    for(int i=x;i<=num+;i+=(i&(-i)))
        q[i]+=y;
}
int get(int x)
{
    int ret=;
    for(int i=x;i;i-=(i&(-i)))
        ret+=q[i];
    return ret;
}
void dfs(int now)
{
    ans+=get(num+)-get(a[now]-);
    gao(b[now],);
    for(int i=h[now];i;i=e[i].nxt)
    {
        dfs(e[i].to);
    }
    gao(b[now],-);
}
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        ans=;
        tot=;
        j=;
        ll p;
        memset(h,,sizeof h);
        memset(fa,,sizeof fa);
        memset(q,,sizeof q);
        scanf("%d%lld",&n,&p);
        rep(i,,n){
            scanf("%lld",&a[i]);
            if(a[i]==)b[i]=1e19;
            else b[i]=p/a[i];
            c[j++]=a[i],c[j++]=b[i];
        }
        sort(c,c+j);
        num=unique(c,c+j)-c;
        rep(i,,n)a[i]=lower_bound(c,c+num,a[i])-c+,b[i]=lower_bound(c,c+num,b[i])-c+;
        rep(i,,n-)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            add(x,y);
            fa[y]=x;
        }
        rep(i,,n)if(!fa[i])dfs(i);
        printf("%lld\n",ans);
    }
    //system("Pause");
    return ;
}