Submission #43258

#	Time	Username	Problem	Language	Result	Execution time	Memory
43258		krauch	Chase (CEOI17_chase)	C++14	100 / 100	1521 ms	361632 KiB

chase

/*
 _    _    _______   _    _
| |  / /  |  _____| | |  / /
| | / /   | |       | | / /
| |/ /    | |_____  | |/ /
| |\ \    |  _____| | |\ \
| | \ \   | |       | | \ \
| |  \ \  | |_____  | |  \ \
|_|   \_\ |_______| |_|   \_\

*/
#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef double ld;
typedef pair <int, int> PII;
typedef pair <ll, ll> PLL;
typedef pair < ll, int > PLI;


#define F first
#define S second
#define pb push_back
#define eb emplace_back
#define right(x) x << 1 | 1
#define left(x) x << 1
#define forn(x, a, b) for (int x = a; x <= b; ++x)
#define for1(x, a, b) for (int x = a; x >= b; --x)
#define mkp make_pair
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define y1 kekekek

#define fname ""

const ll ool = 1e18 + 9;
const int oo = 1e9 + 9, base = 1e9 + 7;
const ld eps = 1e-7;
const int N = 1e5 + 6, M = 111;

int n, K;
ll res[N][2], dd[N][M][2], du[N][M][2], mx[M][2], p[N], ans;
vector < int > g[N];

void dfs(int v, int par) {
    ll sum = 0;

    for (auto to : g[v]) {
        if (to == par) continue;
        dfs(to, v);
        sum += p[to];
    }

    forn(i, 0, K + 1) forn(j, 0, 1) du[v][i][j] = dd[v][i][j] = res[v][j] = mx[i][j] = 0;
    mx[0][1] = -ool;

    for (auto to : g[v]) {
        if (to == par) continue;
        forn(i, 0, K) {
            res[v][0] = max(res[v][0], max(du[to][i][0], du[to][i][1] + p[v]) + max(mx[K - i][0], mx[K - i][1]));
            if (i < K) res[v][1] = max(res[v][1], max(du[to][i][0], du[to][i][1] + p[v]) + sum - p[to] + max(mx[K - i - 1][0], mx[K - i - 1][1]));
            du[v][i][0] = max(du[v][i][0], max(du[to][i][0], du[to][i][1] + p[v]));
            du[v][i + 1][1] = max(du[v][i + 1][1], max(du[to][i][0], du[to][i][1] + p[v]) + sum - p[to]);
            dd[v][i][0] = max(dd[v][i][0], max(dd[to][i][0], dd[to][i][1]));
            dd[v][i + 1][1] = max(dd[v][i + 1][1], max(dd[to][i][0], dd[to][i][1]) + sum);
        }
        forn(i, 0, K) {
            mx[i][0] = max(mx[i][0], dd[to][i][0]);
            mx[i][1] = max(mx[i][1], dd[to][i][1]);
        }
    }

    du[v][0][1] = -ool;
    dd[v][0][1] = -ool;
    du[v][1][1] = max(du[v][1][1], sum);
    dd[v][1][1] = max(dd[v][1][1], sum);
    res[v][1] = max(res[v][1], sum);
    forn(i, 1, K) {
        du[v][i][0] = max(du[v][i][0], du[v][i - 1][0]);
        du[v][i][1] = max(du[v][i][1], du[v][i - 1][1]);
        dd[v][i][0] = max(dd[v][i][0], dd[v][i - 1][0]);
        dd[v][i][1] = max(dd[v][i][1], dd[v][i - 1][1]);
    }

    ans = max(ans, max(res[v][0], res[v][1] + p[par]));
    ans = max(ans, max(du[v][K][0], du[v][K][1] + p[par]));
    ans = max(ans, max(dd[v][K][0], dd[v][K][1] + p[par]));
}

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

	#ifdef krauch
        freopen("input.txt", "r", stdin);
    #else
        //freopen(fname".in", "r", stdin);
        //freopen(fname".out", "w", stdout);
    #endif

    cin >> n >> K;
    forn(i, 1, n) {
        cin >> p[i];
    }

    forn(i, 1, n - 1) {
        int x, y;
        cin >> x >> y;
        g[x].eb(y);
        g[y].eb(x);
    }

    if (!K) {
        cout << "0\n";
        return 0;
    }

    dfs(1, 0);
    forn(i, 1, n) reverse(all(g[i]));
    dfs(1, 0);

    cout << ans << "\n";

	return 0;
}

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