hdu5877 Weak Pair 线段树 （2016 icpc dalian online 1010）



Weak Pair

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 597    Accepted Submission(s): 207

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) 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 nodesu 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

12 31 21 2

 

Sample Output

1

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

题意：有根树，每个节点有非负值，若存在一对（u，v）满足u是v的祖先并且w[u]*w[v]<=k,则该对满足要求，问一共有多少对。

思路：线段树离散化存储所有值和k/w[u]的值，从根开始往下遍历，访问到节点i时，查询该已经出现的节点及其祖先节点小于k/w[i]的个数，这里用线段树维护，线段树存储的值即某个值已经出现的次数。

#include <cstdlib>#include <cstring>#include <algorithm>#include <vector>#include <cmath>#include <cstdio>using namespace std;typedef long long ll;const ll Maxn=1e5+100;int n,numm,head[Maxn],tot,deep[Maxn];ll num[Maxn],ans,k,x[2*Maxn];struct node{    ll maxn;    int l,r;} t[6*Maxn];void pushup(int i){t[i].maxn=t[i<<1|1].maxn+t[i<<1].maxn;}void build(int i,int l,int r){    int mid=(l+r)/2;    t[i].l=l;    t[i].r=r;    if(l==r)    {        t[i].maxn=0;        return;    }    build(2*i,l,mid);    build(2*i+1,mid+1,r);pushup(i);}ll query(int i,int x,int y){    int l=t[i].l;    int r=t[i].r;    int mid=(l+r)/2;    if(t[i].l>=x&&t[i].r<=y)        return t[i].maxn;ll res=0;        if(x<=mid)            res+= query(2*i,x,y);        if(y>mid)            res+= query(2*i+1,x,y);return res;}void Modify(int i,int x,int der){    int l,r,mid;    l = t[i].l;    r = t[i].r;    mid = (l + r) / 2;    if(l == r){        t[i].maxn += der;        return;    }    if(x <= mid)   Modify(2*i,x,der);    else    Modify(2*i+1,x,der);    pushup(i);}struct node1{    int next;    int to;} edge[Maxn];void addedge(int from,int to){    edge[tot].to=to;    edge[tot].next=head[from];    head[from]=tot++;}void dfs(int u,int fa){    int o=lower_bound(x,x+numm,k/num[u])-x;ans+=query(1,0,o);    int kk=lower_bound(x,x+numm,num[u])-x;Modify(1,kk,1);    for(int i=head[u]; ~i; i=edge[i].next)    {        int v=edge[i].to;        dfs(v,u);    }    Modify(1,kk,-1);}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%I64d",&n,&k);        int cnt=0;        for(int i=1; i<=n; i++)        {            scanf("%I64d",&num[i]);            x[cnt++]=num[i];        }        for(int i=1; i<=n; i++)            x[cnt++]=k/num[i];        sort(x,x+cnt);        numm=unique(x,x+cnt)-x;        tot=0;        memset(head,-1,sizeof(head));        memset(deep,0,sizeof(deep));        for(int i=0; i<n-1; i++)        {            int u,v;            scanf("%d %d",&u,&v);            addedge(u,v);            deep[v]++;        }        ans=0;        build(1,0,numm);        for(int i=1; i<=n; i++)            if(deep[i]==0)                dfs(i,-1);        printf("%I64d\n",ans);    }    return 0;}