Similar Pair _ HackerRank, pairhackerrank

Source: Internet
Author: User

Similar Pair _ HackerRank, pairhackerrank

Hacker rank is really more difficult than leetcode ..

This question is a bit clever .. Each path is searched in depth and then enumerated. Most people think about it, but the key is that this definitely times out. The trick is to create a line segment tree for each path to accelerate the query. The complexity of each similar query changes from O (h) to O (lgh )..

Two mistakes were made.

(1) Use long to store the line segment tree and possible similar pairs.

(2) The value minus T may be less than 0, and the value plus T may be greater than n.

import*;import java.util.*;import java.text.*;import java.math.*;public class Solution {public static LinkedList<Integer>[] nodes = new LinkedList[100002];static int n , t, root;    public static void main(String[] args) {        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */        Scanner scan = new Scanner(;        n = scan.nextInt();    t = scan.nextInt();    long[] stree = new long[4*n+1];        for(int i=1;i<=n;i++)    nodes[i] = new LinkedList<Integer>();        int[] idegree = new int[n+1];        for(int i=1;i<n;i++)    {    int par = scan.nextInt();    int chd = scan.nextInt();        nodes[par].addFirst(chd);    idegree[chd]++;    }        for(int i=1;i<=n;i++)    {    if(idegree[i] == 0)    {    root = i;    break;    }    }        long[] pairs = new long[1];        depthSearch(root,stree,pairs);        System.out.println(pairs[0]);        }        public static void depthSearch(int nodeval, long[] stree, long[] pairs){        int min = (nodeval - t < 1) ? 1 : nodeval - t;    int max = (nodeval + t > n) ? n : nodeval + t;        pairs[0] += query(stree,1,1,n,min, max);        updateTree(stree,1,1,n,nodeval,1);        for(int chd : nodes[nodeval]){    depthSearch(chd, stree, pairs);    }        updateTree(stree,1,1,n,nodeval,-1);    }            public static void updateTree(long[] tree, int node,int tl, int tr, int val, long opt){    if(val < tl || val > tr || tl > tr)    return;        tree[node] += opt;        int m = (tl + tr) >> 1;        if(tl == tr)    return;    else if(val <= m)    updateTree(tree,node<<1,tl,m,val,opt);    else    updateTree(tree,node<<1|1,m+1,tr,val,opt);    }           public static long query(long[] tree, int node, int tl, int tr, int min, int max){        if(max < tl || min > tr)    return 0;        else if(max == tr && min == tl)    return tree[node];        else{    int mid = (tl + tr) >> 1;    int lmax = (mid < max) ? mid : max;    int rmin = (min > mid) ? min : mid + 1;    return query(tree,node<<1, tl, mid, min, lmax) + query(tree,node<<1|1, mid+1, tr, rmin, max);    }        }}

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