hdu 5877 线段树（2016 ACM/ICPC Asia Regional Dalian Online）

Weak Pair

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 439 Accepted Submission(s): 155


[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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

[align=left]Output[/align]
For each test case, print a single integer on a single line denoting the number of weak pairs in the tree.

[align=left]Sample Input[/align]

1
2 3
1 2
1 2

[align=left]Sample Output[/align]

1

/*
hdu 5877 线段树

problem:
给你一棵n个节点的有根树,每个节点有价值a[i].   问有多少个点对(u,v)满足:u是v的祖先且a[u]*a[v] <= k

solve:
先找出这个树的根节点.
因为要求u是v的祖先,所以相当于v的父亲到根节点的所有点. 所以可以在树的遍历的时候把走过点的值存入线段树中,当走到第i个节点值
求出线段树中[1,k/a[i]]总共有多少个值就行. 然后递归回退时在把这个点删掉.
数据很大所以再进行一下离散化处理.

hhh-2016-09-11 09:22:59
*/
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <stdio.h>
#include <cstring>
#include <vector>
#include <math.h>
#include <queue>
#include <set>
#define lson  i<<1
#define rson  i<<1|1
#include <map>
#define ll long long

using namespace std;

const int maxn = 200100;

int a[maxn];

struct Node
{
int l,r;
int val;
} tree[maxn <<2];

void push_up(int i)
{
tree[i].val=  tree[lson].val  + tree[rson].val;
}

void build(int i,int l,int r)
{
tree[i].l = l,tree[i].r = r;
tree[i].val = 0;
if(l == r)
{
return ;
}
int mid = (tree[i].l + tree[i].r) >> 1;
build(lson,l,mid);
build(rson,mid+1,r);
push_up(i);
}

void update(int i,int k,int va)
{
if(tree[i].l == tree[i].r && tree[i].l == k)
{
tree[i].val += va;
return;
}
int mid = (tree[i].l + tree[i].r) >> 1;
if(k <= mid)
update(lson,k,va);
else
update(rson,k,va);
push_up(i);
}

int query(int i,int l,int r)
{
if(l > r)
return 0;
if(tree[i].l >= l && tree[i].r <= r)
{
return tree[i].val;
}
int tans = 0;
int mid = (tree[i].l + tree[i].r ) >> 1;
if(l <= mid)
tans += query(lson,l,r);
if(r > mid)
tans += query(rson,l,r);
return tans;
}

struct node
{
int next;
int to;
} edge[maxn];
ll k;
int ta,tot,n;
int head[maxn];
int deep[maxn];
int t[maxn];
void addedge(int from,int to)
{
edge[tot].to=to;
edge[tot].next=head[from];
head[from]=tot++;
}
ll ans = 0;
void dfs(int u,int fa)
{
int o=lower_bound(t,t+ta,k/a[u])-t;
ans+=query(1,0,o);

int tk=lower_bound(t,t+ta,a[u])-t;
update(1,tk,1);
for(int i=head[u]; ~i; i=edge[i].next)
{
int v=edge[i].to;
dfs(v,u);
}
update(1,tk,-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",&a[i]);
t[cnt++]=a[i];
}
for(int i=1; i<=n; i++) t[cnt++]=k/a[i];

sort(t,t+cnt);
ta=unique(t,t+cnt)-t;

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,ta);
for(int i=1; i<=n; i++)
if(deep[i]==0)
dfs(i,-1);
printf("%I64d\n",ans);
}
return 0;
}