Cod sursa(job #2833484)

Utilizator CraiuAndreiCraiu Andrei David CraiuAndrei Data 15 ianuarie 2022 11:48:47
Problema Lowest Common Ancestor Scor 40
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 2.08 kb
#include <bits/stdc++.h>
using namespace std;

ifstream fin("lca.in");
ofstream fout("lca.out");
vector<int> h[100003];
int n, e[1000001], niv[2000001], p, poz[100003];
int rmq[21][1000001], p2[100001];
/**
e - retine nodurile in ordinea parcurgerii Euler
niv[i] = nivelul pe care se afla nodul e[i] , i=1..p
i=1..n, poz[i]=o pozitie unde se gaseste nodul i in e[]

p2[i] = lungimea putere a lui 2 maxima a unui interval care
       are lungimea i
rmq[i][j] = pozitia minimului din intervalul [j, j+2^i-1]
i = 0,1,...k, unde 2^k <= p

1 2 3 4 5 6 7 8 9 10
6 4 2 1 7 5 8 6 3 7

     1 2 3 4 5 6 7 8 9
p2 = 0 1 1 2 2 2 2 3 3
*/

void Euler(int k, int nivel)
{
    e[++p] = k;
    niv[p] = nivel;
    poz[k] = p;
    for (int i : h[k])
    {
        Euler(i, nivel + 1);
        e[++p] = k;
        niv[p] = nivel;
    }

}

int main()
{
    int i, j, m;
    fin >> n >> m;
    for (i = 2; i <= n; i++)
    {
        fin >> j;
        h[j].push_back(i);
    }
    Euler(1, 1);

    /// -------------- RMQ - range minimum query ------
    p2[1] = 0;
    for (i = 2; i <= p; i++)
        p2[i] = 1 + p2[i / 2];
    /// rmq[0][i], i=1..p
    for (i = 1; i <= p; i++)
        rmq[0][i] = i; /// e[i] este minim si se afla la poz i
    /// rmq[i][j], i=1,2,3...
    for (i = 1; (1 << i) <= p; i++)
        for (j = 1; j <= p - (1 << i) + 1; j++)
        {
            int len = (1 << (i - 1));
            /// rmq[i][j] = din rmq[i-1][j] si
            /// rmq[i-1][j + 2^(i-1)]
            rmq[i][j] = rmq[i - 1][j];
            if (niv[rmq[i - 1][j]] > niv[rmq[i - 1][j + len]])
                rmq[i][j] = rmq[i - 1][j + len];
        }
    /// interogarile:
    while (m--)
    {
        fin >> i >> j;
        i = poz[i];
        j = poz[j];
        if (i > j) swap(i, j);
        int expo = p2[j - i + 1];
        int len = 1 << expo;
        int sol1 = rmq[expo][i];
        int sol2 = rmq[expo][j - len + 1];
        if (niv[sol1] > niv[sol2]) fout << e[sol2] << "\n";
        else fout << e[sol1] << "\n";
    }
    fout.close();
    return 0;
}