hdu5877 Weak Pair

Weak Pair

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1556    Accepted Submission(s): 501

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 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

12 31 21 2

 

Sample Output

1

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5876 5875 5874 5873 5872 

题意：给你一棵树，树上每个点都有权值，再给你它的父子关系，问最后在同一棵子树上的任意两个点权值相乘<=k的有多少对。

思路：看别人的做法都是用splay做的，表示我还没学会就用了其他方法做了，先把点权值和k/点权值离散化，然后dfs树，每到一个节点就用线段树查询当前父节点有多少个<=k/该点权值，然后再用线段树单点增加，返回时再单点删除就好了。下面给代码：

#include<iostream>#include<cmath>#include<queue>#include<cstdio>#include<queue>#include<algorithm>#include<cstring>#include<string>#include <functional>#define maxn 100005#define inf 0x3f3f3f3f#define lson l1,mid,l2,r2,now<<1#define rson mid+1,r1,l2,r2,now<<1|1#define ll now<<1#define rr now<<1|1typedef long long LL;using namespace std;int vis[maxn], value[maxn], head[maxn], sumnum[maxn << 3];LL x[maxn << 1];int length1, ans, n, length2;LL k;struct node{int v, next;}p[maxn];void build(int l, int r, int now){if (l == r){sumnum[now] = 0;return;}int mid = (l + r) >> 1;build(l, mid, ll);build(mid + 1, r, rr);sumnum[now] = 0;}void add(int l1, int r1, int num, int now){if (l1 == r1){sumnum[now]++;return;}int mid = (l1 + r1) >> 1;if (num <= mid)add(l1, mid, num, ll);elseadd(mid + 1, r1, num, rr);sumnum[now] = sumnum[ll] + sumnum[rr];}void del(int l1, int r1, int num, int now){if (l1 == r1){sumnum[now]--;return;}int mid = (l1 + r1) >> 1;if (num <= mid)del(l1, mid, num, ll);elsedel(mid + 1, r1, num, rr);sumnum[now] = sumnum[ll] + sumnum[rr];}int query(int l1, int r1, int l2, int r2, int now){if (l2 <= l1&&r2 >= r1){return sumnum[now];}int mid = (l1 + r1) >> 1;int a = 0, b = 0;if (l2 <= mid)a = query(lson);if (r2>mid)b = query(rson);return a + b;}void dfs(int son){int num = lower_bound(x + 1, x + length1 + 1, k / value[son]) - x;ans += query(1, length1, 1, num, 1);int pos = lower_bound(x + 1, x + length1 + 1, value[son]) - x;add(1, length1, pos, 1);for (int i = head[son]; ~i; i = p[i].next){dfs(p[i].v);}del(1, length1, pos, 1);}int main(){int t;    scanf("%d", &t);    while (t--){scanf("%d%lld", &n, &k);memset(head, -1, sizeof(head));memset(vis, 0, sizeof(vis));ans = 0;for (int i = 1; i <= n; i++){scanf("%d", &value[i]);x[i] = value[i];}for (int i = n + 1; i <= (n << 1); i++){x[i] = k / value[i - n];}sort(x + 1, x + (n << 1) + 1);length1 = unique(x + 1, x + (n << 1) + 1) - x;build(1, length1, 1);length2 = 0;for (int i = 0; i<n - 1; i++){int u, v;scanf("%d%d", &u, &v);vis[v] = 1;p[length2].v = v;p[length2].next = head[u];            head[u]=length2++;}for (int i = 1; i <= n; i++){if (!vis[i])dfs(i);}printf("%d\n", ans);}}