Submission #511984

#	Time	Username	Problem	Language	Result	Execution time	Memory
511984		Monarchuwu	Chase (CEOI17_chase)	C++17	100 / 100	447 ms	317284 KiB

chase

#include<iostream>
#include<algorithm>
#include<vector>
#define all(x) x.begin(), x.end()
using namespace std;
typedef long long ll;

const int N = 1e5 + 9, K = 100 + 2;
int n, V;
int p[N];
vector<int> g[N];
ll sum_p[N];

ll dp1[N][K];   // dp1[u][j]: bắt đầu tại u, sẽ dùng j lượt (đi vào subtree)
void dfs_1(int u, int par) {
    for (int v : g[u]) if (v != par) {
        dfs_1(v, u);
        for (int j = 1; j <= V; ++j) {
            dp1[u][j] = max({ dp1[u][j], dp1[v][j], dp1[v][j - 1] + sum_p[v] - p[u] });
        }
    }
}
ll dp2[N][K];   // dp2[u][j]: bắt đầu tại u, sẽ dùng j lượt (đi lên root)
ll tmp[K];
ll prf[N][K], suf[N][K];
void dfs_2(int u, int par) {

    fill(tmp, tmp + V + 1, 0);
    for (int v : g[u]) if (v != par) {
        copy(tmp, tmp + V + 1, prf[v]);
        for (int j = 1; j <= V; ++j) {
            tmp[j] = max({ tmp[j], dp1[v][j], dp1[v][j - 1] + sum_p[v] - p[u] });
        }
    }

    reverse(all(g[u]));
    fill(tmp, tmp + V + 1, 0);
    for (int v : g[u]) if (v != par) {
        copy(tmp, tmp + V + 1, suf[v]);
        for (int j = 1; j <= V; ++j) {
            tmp[j] = max({ tmp[j], dp1[v][j], dp1[v][j - 1] + sum_p[v] - p[u] });
        }
    }
    reverse(all(g[u]));

    for (int v : g[u]) if (v != par) {

        copy(dp2[u], dp2[u] + V + 1, tmp);
        for (int j = 1; j <= V; ++j) {
            tmp[j] = max({ tmp[j], prf[v][j], suf[v][j] });
        }

        for (int j = 1; j <= V; ++j) {
            dp2[v][j] = max({ dp2[v][j], tmp[j], tmp[j - 1] + sum_p[u] - p[v] });
        }

        dfs_2(v, u);
    }
}

int main() {
    cin.tie(NULL)->sync_with_stdio(false);
    cin >> n >> V;
    for (int i = 1; i <= n; ++i) cin >> p[i];
    for (int i = 1, u, v; i < n; ++i) {
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    for (int u = 1; u <= n; ++u)
        for (int v : g[u]) sum_p[u] += p[v];

    dfs_1(1, 0);
    dfs_2(1, 0);
    ll ans(0);
    for (int u = 1; u <= n; ++u) {
        ans = max({ ans, dp1[u][V], dp2[u][V] });
        for (int j = 0; j < V; ++j) {
            ans = max(ans, max(dp1[u][j], dp2[u][j]) + sum_p[u]);
        }
    }
    cout << ans << '\n';
}
/**  /\_/\
 *  (= ._.)
 *  / >0  \>1
**/
/*
==================================================+
INPUT:                                            |
--------------------------------------------------|
12 2
2 3 3 8 1 5 6 7 8 3 5 4
2 1
2 7
3 4
4 7
7 6
5 6
6 8
6 9
7 10
10 11
10 12
--------------------------------------------------|
==================================================+
OUTPUT:                                           |
--------------------------------------------------|
36
--------------------------------------------------|
==================================================+
*/

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 30 / 30
• Subtask 4: 30 / 30