제출 #46969

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
46969		nickyrio	Chase (CEOI17_chase)	C++17	40 / 100	4024 ms	184932 KiB

chase

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define FORD(i, a, b) for (int i = (a); i >= (b); --i)
#define REP(i, a) for (int i = 0; i < (a); ++i)
#define DEBUG(x) { cerr << #x << '=' << x << endl; }
#define Arr(a, l, r) { cerr << #a << " = {"; FOR(_, l, r) cerr << ' ' << a[_]; cerr << "}\n"; }
#define N 100100
#define pp pair<int, int>
#define endl '\n'
#define IO ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define taskname ""
#define bit(S, i) (((S) >> (i)) & 1)
void Max(long long &a, long long b) { a = a < b ? b : a; }
void Min(long long &a, long long b) { a = a < b ? a : b; }
using namespace std;

int n, k;
vector<int> e[N];
long long ans, a[N], DtoU[N][111], UtoD[N][111];

void dfs(int u, int p) {
    long long sum = 0;
    for (int v : e[u]) sum += a[v];
    DtoU[u][1] = UtoD[u][1] = sum;
    /*FOR(i, 1, k) {
        // Choose u
        if (i != 1 && p != -1) Max(UtoD[u][i], UtoD[p][]
    }*/
    for (int v : e[u]) if (v != p) {
        dfs(v, u);
        FOR(i, 1, k) {
            // Choose u
            if (i != 1) Max(DtoU[u][i], DtoU[v][i - 1] + sum - a[v]);
            // Don't choose u
            Max(DtoU[u][i], DtoU[v][i]);
        }
    }
}

int main() {
    #ifdef NERO
    freopen("test.inp","r",stdin);
    freopen("test.out","w",stdout);
    double stime = clock();
    #else 
        //freopen(taskname".inp","r",stdin);
        //freopen(taskname".out","w",stdout);
    #endif //NERO
    IO;
    cin >> n >> k;
    FOR(i, 1, n) cin >> a[i];
    FOR(i, 2, n) {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    if (k == 0) {
        cout << 0;
        return 0;
    }
    ans = 0;
    FOR(root, 1, n) {
        FOR(i, 1, n) FOR(j, 1, k) DtoU[i][j] = 0;
        dfs(root, -1);
        FOR(i, 1, k) Max(ans, DtoU[root][i]);
    }
    cout << ans;
    #ifdef NERO
    double etime = clock();
    cerr << "\nExecution time: " << (etime - stime) / CLOCKS_PER_SEC * 1000 << " ms.\n";
    #endif // NERO
}

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 0 / 30
• Subtask 4: 0 / 30