Submission #59742

#	Time	Username	Problem	Language	Result	Execution time	Memory
59742		Benq	Museum (CEOI17_museum)	C++11	100 / 100	835 ms	204372 KiB

museum

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
 
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
 
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
 
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 10001;
 
int n,k,x;
vpi adj[MX];
vi bes[MX][2];

vi bet(vi a, vi b) {
    F0R(i,sz(a)) a[i] = min(a[i],b[i]);
    return a;
}

vi conv(vi a, vi b) {
    vi res(sz(a)+sz(b)-1);
    F0R(i,sz(res)) res[i] = MOD;
    F0R(i,sz(a)) F0R(j,sz(b)) res[i+j] = min(res[i+j],a[i]+b[j]);
    return res;
}

void comb(int a, int b) {
    bes[a][0] = bet(conv(bes[a][0],bes[b][1]),conv(bes[a][1],bes[b][0]));
    bes[a][1] = conv(bes[a][1],bes[b][1]);
}

void dfs(int a, int b) {
    bes[a][0] = bes[a][1] = {0,0};
    for (auto x: adj[a]) if (x.f != b) {
        dfs(x.f,a);
        FOR(i,1,sz(bes[x.f][0])) {
            bes[x.f][0][i] += x.s;
            bes[x.f][1][i] += 2*x.s;
        }
        comb(a,x.f);
    }
    /*cout << a << " | ";
    for (int i: bes[a][0]) cout << i << " ";
        cout << "\n";*/
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> k >> x;
    F0R(i,n-1) {
        int a,b,c; cin >> a >> b >> c;
        adj[a].pb({b,c}), adj[b].pb({a,c});
    }
    dfs(x,0);
    cout << bes[x][0][k];
}
 
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

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