HDU 5877 Weak Pair

Problem Description

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

一道数据结构好题，从不同的角度有不同的做法，

树上直接做可以自上向下统计每个点对于祖先的贡献，用平衡树或动态线段树可以搞定，离散后可以树状数组或线段树。

自下向上可以用线段树或平衡树的启发式合并，还可以树形转线形，询问区间里小于某个值的数有几个，可以线段树加二分，

或者可持久化线段树等等。这里我用了最近学会的线段树启发式合并自下向上用拓扑排序搞定。

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define ff first
#define ss second
#define mp(i,j) make_pair(i,j)
#define pb push_back
#define pii pair<int,LL>
#define in(x) scanf("%d", &x);
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-4;
const int INF = 0x7FFFFFFF;
const int mod = 998244353;
const int N = 1e5 + 10;
int T, n, x, y, l, r;
LL m;
int fa[N], cnt[N], a[N];
int f[35 * N], L[35 * N], R[35 * N], g[N], tot;

int node()
{
  L[tot] = R[tot] = f[tot] = 0; return tot++;
}

void make(int &x, int l, int r, int u)
{
  if (!x) x = node();
  f[x] = 1;
  if (l == r) return;
  int mid = l + r >> 1;
  if (u <= mid) make(L[x], l, mid, u);
  else make(R[x], mid + 1, r, u);
}

int find(int x, int l, int r, LL u)
{
  if (!x || u < l) return 0;
  if (l == r) return f[x];
  int mid = l + r >> 1;
  if (u <= mid) return find(L[x], l, mid, u);
  return f[L[x]] + find(R[x], mid + 1, r, u);
}

void merge(int &x, int y, int l, int r)
{
  if (!x || !y) { x = x^y; return; }
  f[x] += f[y]; 
  if (l == r) return;
  int mid = l + r >> 1;
  merge(L[x], L[y], l, mid);
  merge(R[x], R[y], mid + 1, r);
}

int main()
{
  scanf("%d", &T);
  while (T--)
  {
    scanf("%d%lld", &n, &m);
    l = 1e9, r = 0, tot = 1;
    rep(i, 1, n)
    {
      scanf("%d", &a[i]); cnt[i] = 0;
      l = min(l, a[i]); r = max(r, a[i]);
    }
    rep(i, 1, n - 1)
    {
      scanf("%d%d", &x, &y);
      cnt[x]++; fa[y] = x;
    }
    queue<int> p;
    LL ans = 0;
    rep(i, 1, n)
    {
      make(g[i] = 0, l, r, a[i]);
      if (!cnt[i]) p.push(i);
      if (1LL * a[i] * a[i] <= m) ans--;
    }
    while (!p.empty())
    {
      int q = p.front(); p.pop();
      ans += find(g[q], l, r, m / a[q]);
      merge(g[fa[q]], g[q], l, r);
      if (!--cnt[fa[q]]) p.push(fa[q]);
    }
    printf("%lld\n", ans);
  }
  return 0;
}