Cod sursa(job #2570724)

Utilizator petrisorvmyVamanu Petru Gabriel petrisorvmy Data 4 martie 2020 18:49:15
Problema Lowest Common Ancestor Scor 100
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 2.02 kb
#include <bits/stdc++.h>
#define FILE_NAME "lca"
#define fast ios_base :: sync_with_stdio(0); cin.tie(0);
#pragma GCC optimize("O3")
#define NMAX 100000 + 100
using namespace std;

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

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;


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

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

int n, t, x, y, ln[2 * NMAX],m;
int lst[2 * NMAX], nivel[2 * NMAX], lvl[2 * NMAX], pd[20][2 * NMAX],euler[2 * NMAX];
vector <int> G[NMAX];


void DFS(int nod)
{
    m++;
    euler[m] = nod;
    nivel[m] = lvl[nod];
    for(auto it : G[nod])
    {
        lvl[it] = lvl[nod] + 1;
        DFS(it);
        euler[++m] = nod;
        nivel[m] = lvl[nod];
    }
    lst[nod] = m;
}
void RMQ()
{
    for(int i = 2; i <= m; ++i)
        ln[i] = 1 + ln[i >> 1];

    for(int i = 1; i <= m; ++i)
        pd[0][i] = i;
    for(int i = 1; i <= ln[m]; ++i)
    {
        int k = 1 << (i - 1);
        for(int j = 1; j <= m; ++j)
            ///pd[i][j] = min(pd[i - 1][j], pd[i - 1][j + k])
            if(nivel[ pd[i-1][j] ] < nivel[ pd[i-1][j + k] ])
                pd[i][j] = pd[i - 1][j];
            else
                pd[i][j] = pd[i - 1][j + k];
    }
}
int LCA(int x, int y)
{
    int a = min(lst[x], lst[y]);
    int b = max(lst[x], lst[y]);
    int d = b - a + 1;
    d = ln[d];
    int ans = pd[d][a];
    if(nivel[ans] > nivel[ pd[d][b - (1 << d) + 1 ] ])
        ans = pd[d][b - (1 << d) + 1 ] ;
    return euler[ans];

}
int main()
{
    f >> n >> t;
    for(int i = 2; i <= n; ++i)
    {
        int x;
        f >> x;
        G[x].push_back(i);
    }
    DFS(1);
    RMQ();
    while(t--)
    {
        int x, y;
        f >> x >> y;
        g << LCA(x,y) << '\n'
;    }
    f.close();
    g.close();
    return 0;
}