Self-Driving Bus


Problem Statement :


Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities.

The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true:

There is a path between every pair of cities which belongs to the subset.
Every city in the path must belong to the subset.

Input Format

The first line contains a single positive integer , n. The n - 1 subsequent lines each contain two positive space-separated integers, ai  and bi , describe an edge connecting two nodes in tree T.

Constraints

1  <=  n  <=  2 x 10^5
1  <=  ai , bi  <=  n 

Output Format

Print a single integer: the number of segments [ L , R ], which are connected in tree T.



Solution :



title-img


                            Solution in C :

In  C++  :





#include<iostream>
#include<vector>
using namespace std;

typedef long long int lli;
typedef pair<lli,lli> ii;
typedef vector<lli> vi;
typedef vector<vi> vii;

const lli MAXN = 200200;
lli par[MAXN], upr[MAXN],dnr[MAXN];
lli run_upr[MAXN], run_dnr[MAXN], grand[MAXN];
vii G(MAXN);

void set_parents(lli m,lli a,lli b, lli p){
  upr[m] = max(m,upr[p]);
  dnr[m] = min(m,dnr[p]);
  par[m] = p;
  grand[m] = grand[p]; 
  for(auto it = G[m].begin() ; it != G[m].end(); it++){
    if((*it) < a || (*it) >= b || (*it) == p) continue;
    set_parents(*it, a, b, m);
  }
}

lli middle_case(lli m, lli a, lli b){
  run_upr[m] = run_dnr[m] = m;
  lli aux;
  for(aux = m ; aux < b && grand[aux] == m ; aux++);
  b = aux;
  for(aux = m ; aux >= a && grand[aux] == m ; aux--);
  a = aux + 1;

  for(lli i = m + 1 ; i < b ; i++){
    run_upr[i] = max(run_upr[i-1], upr[i]);
    run_dnr[i] = min(run_dnr[i-1], dnr[i]);
  }

  for(lli i = m - 1 ; i >= a ; i--){
    run_upr[i] = max(run_upr[i+1], upr[i]);
    run_dnr[i] = min(run_dnr[i+1], dnr[i]);
  }
  lli total = 0; // {m}
  // Contamos [i,d] con i <= m/2, d>= m/2
  for(lli d = m, l = m + 1, r = m+ 1, ct = 0; d < b ; d++){
    if(d != run_upr[d]) continue;
    for(; l - 1>= a && d >= run_upr[l- 1] ;l--){
      if(l-1 == run_dnr[l-1]) ct++;
    }
    for(; r - 1> run_dnr[d] && r > l; r--){
      if(r - 1 == run_dnr[r-1]) ct--;
    }
    total += ct;
  }
  return total;
}

lli solve(lli a, lli b){
  if(a == b){
    return 0;
  }
  if(a + 1 == b){
    return 1;
  }
  lli m = (a+b)/2;
  lli x,y,z;
  upr[m] = par[m] = dnr[m] = grand[m] = m;
  set_parents(m,a,b,m);
  x = middle_case(m,a,b);
  y = solve(a,m);
  z = solve(m + 1,b);
  return (x+y+z);
}

int main(){
  lli n, a,b;
  cin >> n;
  for(int i = 1 ; i<n;i++){
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  cout << solve(0,n) << endl;
  return 0;
}








In   Java :








import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
  
        Scanner sc = new Scanner(System.in);        
        int nodeNum = Integer.parseInt(sc.nextLine());
List<List<Integer>> refer = new ArrayList<List<Integer>>();
        for(int i=0; i<nodeNum; i++) {
            refer.add(new ArrayList<Integer>());
        }
        
        while(sc.hasNextLine()) {
            String[] pair = sc.nextLine().split(" ");
            putRefer(refer, pair[0], pair[1]);
        }
        
        int result = 0;
        for(int i=1; i<=nodeNum; i++) {
            boolean[] battleFront = new boolean[nodeNum];
            expandFront(i, battleFront, refer);
            result++;
            int upto = i;
            for(int j=i+1; j<=nodeNum; j++) {
                if(battleFront[j-1]) {
                    expandFront(j, battleFront, refer);
                    if(allInFront(battleFront, upto, j)) {
                        result++;
                        for(int k=upto+1; k<j; k++) {
                             expandFront(k, battleFront, refer);
                        }
                        upto = j;
                    }
                }
            }
        }
        
        System.out.println(result);
    }
    
    public static void putRefer(List<List<Integer>> refer, String s1, String s2) {
        int a = Integer.parseInt(s1);
        int b = Integer.parseInt(s2);
        List<Integer> tmp = refer.get(a-1);
        tmp.add(b);
        tmp = refer.get(b-1);
        tmp.add(a);
    }
    
    public static void expandFront(int i, 
boolean[] battleFront, List<List<Integer>> refer) {
        List<Integer> tmp = refer.get(i-1);
        battleFront[i-1] = true;
        if(tmp == null) {
            return;
        }
        for(Integer thisInt : tmp) {
            battleFront[thisInt-1] = true;
        }
    }
    
    public static boolean allInFront(boolean[] battleFront, int upto, int j) {
        for(int i=upto; i<j-1; i++) {
            if(battleFront[i] == false) {
                return false;
            }
        }
        return true;
    }
    
    public static void printArray(boolean[] battleFront) {
        for(boolean b : battleFront) {
            System.out.print(b + " ");
        }
        System.out.println(" ");
    }
    
    public static void printList(List<List<Integer>> refer) {
        for(List<Integer> l : refer) {
            if(l == null) {
                System.out.println("null ");
            }
            else {
                for(Integer ii : l) {
                    System.out.print(ii + " ");                    
                }
                System.out.println(" ");
            }
        }        
    }
}









In   Python3 :








from heapq import *
n=int(input())
neighbors = {}

for x in range(n):
    neighbors[x] = []
for i in range(n-1):
    a, b = map(int,input().split())
    neighbors[a-1].append(b-1)
    neighbors[b-1].append(a-1)
def search(source):
    ans = 0
    cur_max = 0
    cur_len = 0
    heap = [source]
    vis = [False for i in range(n)]
    while len(heap) > 0:
        x = heappop(heap)
        cur_max = max(cur_max, x)
        cur_len += 1
        vis[x] = True
        if cur_max - source + 1 == cur_len:
            ans += 1
        for y in neighbors[x]:
            if y >= source and vis[y] == False:
                heappush(heap, y)
    return ans
ans = 0
prev = 0
for x in range(n-1, -1, -1):
    neigh = 0
    plus = 0
    for y in neighbors[x]:
        if y > x:
            neigh += 1
        if y == x + 1:
            plus = 1
    if plus == neigh and plus == 1:
        prev += 1
    else:
        prev = search(x)
    ans += prev
print(ans)
                        








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