Weak Pair HDU - 5877(树状数组+离散化）

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, NN and kk, 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 uu and vv , where node uu is the parent of node vv.

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
犯了一个超级超级低级错误，数组开小了10倍，一直和我说超时，实在没办法就重新写了一遍才发现，wa到怀疑人生GG

见上一个同类型的题目

#include
#include
#include
#include
#include
#include
#include
#define N 100005
using namespace std;
typedef long long ll;
int n;
ll k;
ll a[N];
ll id[200005];
int cnt,num;
bool inD[N];
ll ans;
int getIndex(ll x)//手写了一个查找离散坐标的函数
{
    int l=1,r=num;
    int mid;
    while(l<=r)
    {
        mid=(r+l)>>1;
        if(id[mid]==x)
            return mid;
        if(id[mid]1;
        else
            r=mid-1;
    }
}
vector<int> graph[N];
int c[200005];
int lowBit(int x)
{
    return x&-x;
}
int sum(int x)
{
    int ans=0;
    while(x>0)
    {
        ans+=c[x];
        x-=lowBit(x);
    }
    return ans;
}
void change(int x,int p)
{
    while(x<=num)
    {
        c[x]+=p;
        x+=lowBit(x);
    }
}//树状数组的三个操作
void dfs(int x)
{
    int index=getIndex(a[x]);
    if(!a[x])
        ans+=sum(num);
    else
        ans+=sum(getIndex(k/a[x]));
    change(index,1);
    for(int i=0;i1);//消除多兄弟节点的影响
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%lld",&n,&k);
        cnt=0;
        ans=0;
        memset(inD,false,sizeof(inD));
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
            graph[i].clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%d",a+i);
            id[++cnt]=a[i];
            if(a[i])
                id[++cnt]=k/a[i];//离散处理 
        }
        sort(id+1,id+cnt+1);
        num=1;
        for(int i=2;i<=cnt;i++)//去重
        {
            if(id[num]!=id[i])
                id[++num]=id[i];
        }
        int x,y;
        for(int i=1;iscanf("%d%d",&x,&y);
            inD[y]=true;
            graph[x].push_back(y);
        }
        for(int i=1;i<=n;i++)
        {
            if(!inD[i])//找到root节点
            {
                dfs(i);
                break;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}