Submission #988690

#	Time	Username	Problem	Language	Result	Execution time	Memory
988690		LOLOLO	Museum (CEOI17_museum)	C++17	80 / 100	3091 ms	785752 KiB

museum

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define           f     first
#define           s     second
#define           pb    push_back
#define           ep    emplace
#define           eb    emplace_back
#define           lb    lower_bound
#define           ub    upper_bound
#define       all(x)    x.begin(), x.end()
#define      rall(x)    x.rbegin(), x.rend()
#define   uniquev(v)    sort(all(v)), (v).resize(unique(all(v)) - (v).begin())
#define     mem(f,x)    memset(f , x , sizeof(f))
#define        sz(x)    (int)(x).size()
#define  __lcm(a, b)    (1ll * ((a) / __gcd((a), (b))) * (b))
#define          mxx    *max_element
#define          mnn    *min_element
#define    cntbit(x)    (ll)__builtin_popcountll(x)
#define       len(x)    (int)(x.length())
 
const int N = 1e4 + 10;
int dp[N][N][2], sz[N];
vector <pair <int, int>> ed[N];
int n, k;

void minimize(int &a, int b) {
    if (a > b) {
        a = b;
    }
}

void dfs(int u, int v) {
    sz[u] = 1;
    dp[u][0][1] = dp[u][1][0] = dp[u][0][0] = dp[u][1][1] = 0;
    for (auto x : ed[u]) {
        if (x.f == v)
            continue;

        dfs(x.f, u);
        for (int a = min(k, sz[u] + sz[x.f]); a >= 1; a--) {
            for (int b = 0; b <= min(sz[x.f], a); b++) {
                minimize(dp[u][a][0], dp[x.f][b][0] + dp[u][a - b][0] + x.s * 2);
                minimize(dp[u][a][1], dp[x.f][b][1] + dp[u][a - b][0] + x.s);
                minimize(dp[u][a][1], dp[x.f][b][0] + dp[u][a - b][1] + x.s * 2);
            }
        }

        sz[u] += sz[x.f];
    }
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    mem(dp, 0x3f);
    int x;
    cin >> n >> k >> x;

    for (int i = 1; i < n; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        ed[a].pb({b, c});
        ed[b].pb({a, c});
    }

    dfs(x, x);

    cout << min(dp[x][k][1], dp[x][k][0]) << '\n';
    return 0;
}

/*
11 8 3
1 3 3
3 2 5
6 4 5
1 11 3
9 1 2
9 10 2
3 7 10
6 7 1
7 8 1
7 5 1
*/

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