제출 #729486

#제출 시각아이디문제언어결과실행 시간메모리
729486MohammadAghilCat Exercise (JOI23_ho_t4)C++17
100 / 100
384 ms40044 KiB
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx2")
#define rep(i,l,r) for(int i = (l); i < (r); i++)
#define per(i,r,l) for(int i = (r); i >= (l); i--)
#define all(x) begin(x), end(x)
#define sz(x) (int)size(x)
#define pb push_back
#define ff first
#define ss second
typedef long long ll;
typedef pair<int, int> pp;

void dbg(){
     cerr << endl;
}
template<typename H, typename... T> void dbg(H h, T... t){
     cerr << h << ", ";
     dbg(t...);
}

void IOS(){
     cin.tie(0) -> sync_with_stdio(0);
     #ifndef ONLINE_JUDGE
          // freopen("inp.txt", "r", stdin);
          // freopen("out.txt", "w", stdout);
          #define er(...) cerr << __LINE__ << " <" << #__VA_ARGS__ << ">: ", dbg(__VA_ARGS__)
     #else
          #define er(...) 0
     #endif
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll mod = 1e9 + 7, maxn = 2e5 + 5, maxk = 31, lg = 19, inf = ll(1e18) + 5;

int par[maxn][lg], h[maxn];
vector<int> adj[maxn];

void dfs(int r, int p){
     par[r][0] = p;
     rep(j,1,lg){
          par[r][j] = par[par[r][j-1]][j-1];
     }
     for(int c: adj[r]) if(c - p){
          h[c] = h[r] + 1;
          dfs(c, r);
     }
}

int lca(int u, int v){
     if(h[u] < h[v]) swap(u, v);
     per(i,lg-1,0) if(h[par[u][i]] >= h[v]) u = par[u][i];
     if(u == v) return u;
     per(i,lg-1,0) if(par[u][i] - par[v][i]) u = par[u][i], v = par[v][i];
     return par[u][0];
}

int dist(int u, int v){
     return h[u] + h[v] - 2*h[lca(u, v)];
}

int pr[maxn];
int get(int x){
     return pr[x] == -1? x: pr[x] = get(pr[x]);
}

ll dp[maxn];

int main(){ IOS();
     int n; cin >> n;
     vector<int> h(n);
     rep(i,0,n){
          cin >> h[i];
          pr[i] = -1;
     }
     rep(i,1,n){
          int u, v; cin >> u >> v; u--, v--;
          adj[u].pb(v), adj[v].pb(u);
     }
     dfs(0, 0);
     vector<int> srt(n); iota(all(srt), 0);
     sort(all(srt), [&](int u, int v){
          return h[u] < h[v];
     });
     for(int r: srt){
          for(int c: adj[r]) if(h[c] < h[r]){ c = get(c);
               dp[r] = max(dp[r], dp[c] + dist(c, r));
               pr[c] = r;
          }
     }
     cout << dp[srt.back()] << '\n';
     return 0;
}
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