Weak Pair(树状数组+离散化)

Weak Pair

http://acm.hdu.edu.cn/showproblem.php?pid=5877

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) ${a_u×a_v}\leq{k}$.

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 $a_1$ to $a_N$.
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:

$1≤N≤10^{5}$

$0≤a_i≤10^{9}$

$0≤k≤10^{18}$

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

题目：

给你一棵有$N$个节点的树,并且每个节点有对应的一个值，现在定义一个叫WEAK PAIR的东西(u,v),当满足一下条件是(u,v)就是WEAK PAIR：
1.u是v的祖先
2.${a_u×a_v}\leq{k}$
然后求树里有多少个WEAK PAIR

题解：

这题用了经典题目求逆序数的思想，用树状数组来维护比当前数要小的数的个数，然后用dfs遍历一遍就能把问题解决。
但应该要注意的是，由于k是$10^{18}$所以数组是不可能存的下的，因此要进行离散化操作就是通过结构体把节点分成值(val)和编号(id)，再按值进行排序，然后映射到一个新的数组tr[]上，这里满足tr[a[i].id]=i(这里的i是指排序之后的第几个)，但要注意的就是里面的值可能是一样的，因此相同的要映射到同一个下标。

补充：

由于样例的数据太弱这里给出我编的数据：
input

3
9 10
2 4 7 9 2 11 7 4 1
1 2
1 3
2 4
2 5
3 6
3 7
3 8
6 9
9 10
2 4 7 9 2 11 7 4 1
1 2
1 3
2 4
2 5
3 6
3 7
3 8
6 9
8 10
2 4 7 9 2 11 7 4
1 2
1 3
2 4
2 5
3 6
3 7
3 8

output

6
6
4

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+5;
int vis[maxn];
int Ans[maxn<<1];
int tr[maxn<<1];
long long ans;
int n;
vector<int>C[maxn<<1];
struct node
{
    int id;
    long long val;
}a[maxn];
int lowbits(int i)
{
    int temp=i&(-i);
//    cout<<i<<" "<<temp<<endl;
    return temp;
}
void add(int x,int k)
{
    while(x<=2*n)
    {
        Ans[x]+=k;
        x+=lowbits(x);
//        cout<<"-------"<<endl; 
    }
}
int getSum(int x)
{
    int ans=0;
    for(int i=x;i>0;i-=lowbits(i))
    {
        ans+=Ans[i];
    }
    return ans;
}
//void check()
//{
//	for(int i=1;i<=2*n;i++)
//	{
//		cout<<setw(5)<<i;
//	}
//	cout<<endl;
//	for(int i=1;i<=2*n;i++)
//	{
//		cout<<setw(5)<<Ans[i];
//	}
//	cout<<endl;
//}
void dfs(int u)
{
    ans+=getSum(tr[u+n]);
        add(tr[u],1);
//			check();
//    cout<<"u="<<u<<" ans="<<ans<<endl;
    int len=C[u].size(),v;
    for(int i=0;i<len;i++)
    {
//        cout<<"add__"<<endl;
        dfs(C[u][i]);
//        cout<<"del__"<<endl;
    }
        add(tr[u],-1);
//        	check();
}
bool cmp(const node& a,const node& b)
{
    if(a.val==b.val)
        return a.id<b.id;
    else
        return a.val<b.val;
}
int main()
{
    int T,i,j,u,v;
    long long k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%lld",&n,&k);
        ans=0;
        memset(vis,0,sizeof(vis));  
        memset(tr,0,sizeof(tr)); 
        memset(a,0,sizeof(a)); 
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&a[i].val);
            a[i+n].val=k/a[i].val;
            a[i].id=i;
            a[i+n].id=i+n;
            C[i].clear();
        }
        sort(a+1,a+1+2*n,cmp);
        int temp=1;
        for(i=1;i<=2*n;i++)
        {
            tr[a[i].id]=temp;
            if(a[i].val!=a[i+1].val)
            {
            	temp++;
			 } 
        }
//        for(i=1;i<=2*n;i++)
//        {
//        	cout<<tr[i]<<" ";
//		}cout<<endl;
        for(i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            C[u].push_back(v);
            vis[v]++;
        }
        int s;
        for(i=1;i<=n;i++)
        {
            if(!vis[i])
            {
                s=i;
                break;
            }
        }
//        add(tr[s],1);
        dfs(s);
//        add(tr[s],-1);
        cout<<ans<<endl;
//        cout<<"--------"<<endl;
    }
}