Weak Pair

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2081    Accepted Submission(s): 643

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 weakif
  (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

思路：将公式au*av<=k变换为 au<=k/av。 在遍历结点v的过程中，统计au<=k/av的节点u的个数。

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN=100005;
int n;
LL k;
vector<int> arc[MAXN];
LL val[MAXN];
LL buf[MAXN+MAXN];
int top;
int deg[MAXN];

int bit[MAXN+MAXN];
void add(int i,int x)
{
    int limit=MAXN+MAXN;
    while(i<limit)
    {
        bit[i]+=x;
        i+=(i&(-i));
    }
}
int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=bit[i];
        i-=(i&(-i));
    }
    return s;
}
int vis[MAXN];
LL res;
void dfs(int u)
{
    vis[u]=1;
    int has1=lower_bound(buf,buf+top,k/val[u])-buf+1;
    int s=sum(has1);
    for(int i=0;i<arc[u].size();i++)
    {
        int to=arc[u][i];
        if(!vis[to])
        {
            dfs(to);
        }
    }
    res+=(sum(has1)-s);
    int has2=lower_bound(buf,buf+top,val[u])-buf+1;
    add(has2,1);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        res=0;
        memset(vis,0,sizeof(vis));
        memset(bit,0,sizeof(bit));
        top=0;
        memset(deg,0,sizeof(deg));
        scanf("%d%lld",&n,&k);
        for(int i=1;i<=n;i++)
        {
            arc[i].clear();
            scanf("%lld",&val[i]);
        }
        for(int i=0;i<n-1;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            arc[u].push_back(v);
            arc[v].push_back(u);
            deg[v]++;
        }
        for(int i=1;i<=n;i++)
        {
            buf[top++]=val[i];
        }
        for(int i=1;i<=n;i++)
        {
            buf[top++]=k/val[i];
        }
        sort(buf,buf+top);
        for(int i=1;i<=n;i++)
        {
            if(deg[i]==0)
            {
                dfs(i);
                break;
            }
        }
        printf("%lld\n",res);
    }
    return 0;
}