Submission #978057

#	Time	Username	Problem	Language	Result	Execution time	Memory
978057		stefanopulos	Nestabilnost (COI23_nestabilnost)	C++17	7 / 100	1566 ms	40272 KiB

code1

#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 ll inf = 1e15 + 5;
const int mxN = 300005; 

int n;
int A[mxN];
int f[mxN];

vector<int> g[mxN];

ll dp[mxN];
ll nula[mxN];
ll get(int v, int p, int k){

    vector<int> q, r; ll sum = 0;
    for(auto u : g[v]){
        if(u != p){
            
            if(A[v] + 1 == A[u]){
                q.pb(u);
            }
            else if(A[u] == 0){
                r.pb(u);
            }
            else{
                sum += dp[u];
            }

        }
    }

    if(A[v] == k - 1){
        
        for(auto u : r){
            sum += min(dp[u], nula[u]); 
        }

        for(auto u : q){
            sum += dp[u];
        }

    }
    else{
    
        for(auto u : q){
            sum += min(dp[u], get(u, v, k)); 
        }

        for(auto u : r){
            sum += dp[u];
        }

    }

    return sum;

}

ll dfs2(int v, int p, int k){

    vector<int> q, r; ll sum = 0;
    for(auto u : g[v]){
        if(u != p){
            
            if(A[v] + 1 == A[u]){
                q.pb(u);
            }
            else if(A[u] == 0){
                r.pb(u);
            }
            else{
                sum += dp[u];
            }

        }
    }

    if(A[v] == k - 1){

        for(auto u : r){
            sum += min(dp[u], nula[u]);
        }   

        for(auto u : q){
            sum += dp[u];
        }

    }
    else{

        for(auto u : q){
            sum += min(dp[u], dfs2(u, v, k)); 
        }

        for(auto u : r){
            sum += dp[u];
        }

    }

    return sum;

}

void dfs(int v, int p){

    for(auto u : g[v]){
        if(u != p){
            dfs(u, v);
        }
    }

    if(A[v] == 0 && p != -1){
        nula[v] = dfs2(v, p, A[p] + 1);
    }

    dp[v] = inf;
    ff(k,A[v] + 1,n)dp[v] = min(dp[v], get(v, p, k) + f[k]);

}

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

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

    dfs(1, -1);

    cout << dp[1] << '\n';



    return 0;
}
/*

7
2 3 0 3 2 0 0
6 8 2 9 9 9 9
1 2
2 3
1 4
4 5
5 6
5 7


7
2 3 0 3 2 0 0
6 8 2 9 9 9 1
1 2
2 3
1 4
4 5
5 6
5 7

// probati bojenje sahovski
*/
 
 
 
 
 

• Subtask 1: 0 / 12
• Subtask 2: 0 / 20
• Subtask 3: 7 / 7
• Subtask 4: 0 / 22
• Subtask 5: 0 / 39