Submission #1094188

#	Time	Username	Problem	Language	Result	Execution time	Memory
1094188		stefanopulos	Cat Exercise (JOI23_ho_t4)	C++17	100 / 100	208 ms	48064 KiB

Main

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ldb;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ldb,ldb> pdd;

#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for(auto& a : x)
 
#define sz(a) (int)(a).size()
#define fi first
#define se second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

const int mod = 1000000007;
const int inf = 1e9 + 5;
const int mxN = 200005; 

int n;
int niz[mxN];

vector<int> g[mxN];

int poz[mxN];

int d[mxN];
int deda[mxN][20];
void dfs(int v, int p){
    deda[v][0] = p;
    ff(i,1,19)deda[v][i] = deda[deda[v][i - 1]][i - 1];
    
    for(auto u : g[v]){
        if(u != p){
            d[u] = d[v] + 1;
            dfs(u, v);
        }
    }

}

int LCA(int x, int y){
    if(d[x] < d[y])swap(x, y);
    fb(i,19,0)if((d[x] - d[y]) & (1 << i))x = deda[x][i];
    fb(i,19,0)if(deda[x][i] != deda[y][i])x = deda[x][i], y = deda[y][i];
    return (x == y ? x : deda[x][0]);
}

int dist(int x, int y){
    return d[x] + d[y] - 2 * d[LCA(x, y)];
}

ll sum[mxN];
int par[mxN];
int findpar(int x){
    return (x == par[x] ? x : par[x] = findpar(par[x]));
}
void unite(int x, int y){
    x = findpar(x), y = findpar(y);
    par[y] = x;
}

int main(){
    cin.tie(0)->sync_with_stdio(0);

    cin >> n;
    ff(i,1,n)cin >> niz[i], poz[niz[i]] = i;
    ff(i,1,n - 1){
        int u, v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }

    dfs(1, 0);

    ff(i,1,n)par[i] = i;

    ff(i,1,n){
        int v = poz[i];

        for(auto c : g[v]){
            if(niz[c] < niz[v]){
                int a = findpar(c);

                sum[v] = max(sum[v], sum[a] + dist(a, v));
                unite(v, a);

            }
        }

    }

    cout << sum[poz[n]] << '\n';

    return 0;
}
/*

5
5 1 2 3 4
1 2
2 3
3 4
4 5

// probati bojenje sahovski
*/
 
 
 
 
 

• Subtask 1: 7 / 7
• Subtask 2: 7 / 7
• Subtask 3: 7 / 7
• Subtask 4: 10 / 10
• Subtask 5: 20 / 20
• Subtask 6: 23 / 23
• Subtask 7: 26 / 26