HDU5877-Weak Pair

Weak Pair

                                                                              Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
                                                                                                         Total Submission(s): 3320    Accepted Submission(s): 1016

Problem Description

You are given a  rooted  tree of  N  nodes, labeled from 1 to  N . To the  i th 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

 

Recommend

wange2014

 

题意：给出一棵树，告诉你树上节点的权值，问有多少对（u，v）满足u是v的祖先，且两个节点权值的乘积小于等于k

解题思路：树状数组+离散化

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int vis[100005 * 2],n;
LL a[100005 * 2], b[100005 * 2], ans, k, xx[100005 * 2];
vector<int>g[100005];

LL lowbit(LL x)
{
	return x&(-x);
}

void add(LL x, int val)
{
	while (x <= 2 * n)
	{
		xx[x] += val;
		x += lowbit(x);
	}
}

LL getsum(LL x)
{
	LL sum = 0;
	while (x>0)
	{
		sum += xx[x];
		x -= lowbit(x);
	}
	return sum;
}

void dfs(int s)
{
	int Size = g[s].size();
	LL sum = getsum(a[n + s]);
	ans += sum;
	add(a[s], 1);
	for (int i = 0; i<Size; i++)
	{
		int v = g[s][i];
		dfs(v);
	}
	add(a[s], -1);
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		ans = 0;
		memset(vis, 0, sizeof(vis));
		memset(xx, 0, sizeof(xx));
		for (int i = 0; i <= n; i++) g[i].clear();
		scanf("%d%lld", &n, &k);
		int u, v;
		for (int i = 1; i <= n; i++)
		{
			scanf("%lld", &a[i]);
			a[i + n] = k / a[i];
			b[i] = a[i], b[i + n] = a[i + n];
		}
		sort(b + 1, b + 2 * n + 1);
		int m = unique(b + 1, b + 2 * n + 1) - b;
		for (int i = 1; i <= 2 * n; i++) a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
		for (int i = 0; i<n - 1; i++)
		{
			scanf("%d%d", &u, &v);
			g[u].push_back(v);
			vis[v]++;
		}
		int s;
		for (int i = 1; i <= n; i++)
			if (vis[i] == 0) { s = i; break; }
		dfs(s);
		printf("%lld\n", ans);
	}
	return 0;
}