Cod sursa(job #2705761)

Utilizator petrisorvmyVamanu Petru Gabriel petrisorvmy Data 13 februarie 2021 11:42:38
Problema Lowest Common Ancestor Scor 30
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 2 kb
#include <bits/stdc++.h>
#define fast ios_base :: sync_with_stdio(0); cin.tie(0);
#pragma GCC optimize("O3")
#define FILE_NAME "lca"
#define NMAX 100000 + 100
#define vs nod<<1
#define vd nod<<1|1
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pi;
typedef pair<ll,ll> llp;
typedef pair<ld,ld> pct;

const ll inf = 1LL << 60;
const ll mod = 1e9 + 7;
const ld eps = 1e-9;

ifstream f(FILE_NAME".in");
ofstream g(FILE_NAME".out");

inline void add(ll &a , ll b)
{
    a += b;
    a %= mod;
}

inline void sub(ll &a, ll b)
{
    a = (a - b + mod) % mod;
}

int n, q, t[NMAX], m, euler[2 * NMAX], lvl[2 * NMAX], lst[2 * NMAX], lg[2 * NMAX];
vector <int> G[NMAX];
int pd[20][NMAX];

void DFS(int nod, int niv)
{
    m++;
    euler[m] = nod;
    lvl[m] = niv;

    for(auto it : G[nod])
    {
        DFS(it, niv + 1);
        m++;
        euler[m] = nod;
        lvl[m] = niv;
    }
    lst[nod] = m;
}

void RMQ()
{
    for(int i = 2; i <= m; ++i)
        lg[i] = 1 + lg[i / 2];
    for(int i = 1; i <= m; ++i)
        pd[0][i] = i;

    for(int i = 1; i <= lg[m]; ++i)
        for(int j = 1; j  <= m; ++j)
        {
            int w = 1 << (i - 1);
            if(lvl[pd[i - 1][j]] < lvl[pd[i - 1][j + w]])
                pd[i][j] =  pd[i - 1][j];
            else
                pd[i][j] =  pd[i - 1][j + w];
        }

}

int LCA(int x, int y)
{
    int a = min(lst[x], lst[y]),
        b = max(lst[x], lst[y]);
    int L = b - a + 1;
    L = lg[L];
    int ans = pd[L][a];
    if(lvl[ans] > lvl[ pd[L][b - (1 << L) + 1] ])
        ans = pd[L][b - (1 << L) + 1];
    return euler[ ans ];
}

int main()
{
    ///fast;
    f >> n >> q;
    for(int i = 2; i <= n; ++i)
    {
        int x;
        f >> x;
        G[x].push_back(i);
    }
    DFS(1,0);
    RMQ();
    while(q--)
    {
        int x, y;
        f >> x >> y;
        g << LCA(x,y) << '\n';
    }
    f.close();
    g.close();
    return 0;
}