Submission #743888

#	Time	Username	Problem	Language	Result	Execution time	Memory
743888		maomao90	Wells (CEOI21_wells)	C++17	0 / 100	31 ms	35540 KiB

wells

// Hallelujah, praise the one who set me free
// Hallelujah, death has lost its grip on me
// You have broken every chain, There's salvation in your name
// Jesus Christ, my living hope
#include <bits/stdc++.h> 
using namespace std;

#define REP(i, s, e) for (int i = (s); i < (e); i++)
#define RREP(i, s, e) for (int i = (s); i >= (e); i--)
template <class T>
inline bool mnto(T& a, T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T& a, T b) {return a < b ? a = b, 1: 0;}
typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> iii;
#define ALL(_a) _a.begin(), _a.end()
#define SZ(_a) (int) _a.size()
#define pb push_back
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ii> vii;
typedef vector<iii> viii;

#ifndef DEBUG
#define cerr if (0) cerr
#endif

const int INF = 1000000005;
const ll LINF = 1000000000000000005ll;
const int MAXN = 1500005;

int n, k;
vi adj[MAXN];

int d[MAXN], mxd[MAXN];
bool pos;
bool dfs(int u, int p) {
    mxd[u] = d[u];
    bool res = d[u] % k == 0;
    int cnt = 0;
    int mx = -INF, smx = -INF;
    for (int v : adj[u]) {
        if (v == p) {
            continue;
        }
        d[v] = d[u] + 1;
        bool tmp = dfs(v, u);
        if (!pos) {
            return 0;
        }
        res |= tmp;
        cnt += tmp;
        mxto(mxd[u], mxd[v]);
        int x = mxd[v] - d[u];
        if (tmp) {
            x = k - d[u] % k - 1;
        }
        if (x >= mx) {
            smx = mx;
            mx = x;
        } else {
            mxto(smx, x);
        }
    }
    if (cnt >= 2 && 2 * d[u] % k != 0) {
        pos = 0;
        return 0;
    }
    if (d[u] % k != 0 && mx + smx + 1 >= k) {
        pos = 0;
        return 0;
    }
    return res;
}

int main() {
#ifndef DEBUG
    ios::sync_with_stdio(0), cin.tie(0);
#endif
    cin >> n >> k;
    REP (i, 1, n) {
        int u, v; cin >> u >> v;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    REP (i, 1, n + 1) {
        d[i] = 0;
        pos = 1;
        dfs(i, -1);
        /*
        if (pos) {
            cerr << i << '\n';
            cout << "YES\n";
            cout << 0 << '\n';
            return 0;
        }
        */
    }
    cout << "NO\n";
    cout << 0 << '\n';
    return 0;
}

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