Submission #1198167

#	Time	Username	Problem	Language	Result	Execution time	Memory
1198167		walizamanee	Museum (CEOI17_museum)	C++20	80 / 100	3086 ms	784412 KiB

museum

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using indexed_set = tree<
    T,
    null_type,
    less<T>,
    rb_tree_tag,
    tree_order_statistics_node_update
>;
 // indexed_set<int> s;
// and use s.insert(x), s.erase(x),
 // *s.find_by_order(k), s.order_of_key(x), etc.
using ll = long long;
vector<pair<int ,unsigned int>> adj[10001];
int n , k , x , ek , dui , siz[10001];
unsigned int one[10001][10001] , two[10001][10001] , c , uno , dos;

void getans( int me , int par ) {
    siz[me] = 1;
    one[me][1] = 0;
    two[me][1] = 0;
    for( int z = 0; z < (int)adj[me].size(); z++ ) {
        if( adj[me][z].first != par ) {
            getans( adj[me][z].first , me );
            siz[me] += siz[adj[me][z].first];
            for( int zz = min( k , siz[me]); zz > 1; zz-- ) {
                for( int y = min( zz - 1 , siz[adj[me][z].first]) ; y > 0; y-- ) {
                  
                            c = one[me][zz - y] + one[adj[me][z].first][y] + 
                            ((ll)2 * adj[me][z].second);
                            one[me][zz] = min( one[me][zz] , c );
                            c = min( two[me][zz - y] + one[adj[me][z].first][y] + 
                            (2 * adj[me][z].second) , one[me][zz - y] + two[adj[me][z].first][y] + adj[me][z].second );
                            two[me][zz] = min( two[me][zz] ,  c );
                        
                    
                }
            }
        }
    }
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    cin >> n >> k >> x;
    for( int z = 1; z < n; z++ ) {
        cin >> ek >> dui >> c;
        adj[ek].push_back({dui , c});
        adj[dui].push_back({ek , c});
    }
    for( int z = 0; z <= n; z++ ) {
        for( int y = 0; y <= n; y++ ) {
            one[z][y] = 200000000;
            two[z][y] = 200000000;
        }
    }
    getans( x , 0 );
    cout << min( one[x][k] , two[x][k] ) << "\n";
    return 0;
}

• Subtask 1: 20 / 20
• Subtask 2: 25 / 25
• Subtask 3: 35 / 35
• Subtask 4: 0 / 20