Submission #405775

#	Time	Username	Problem	Language	Result	Execution time	Memory
405775		534351	The Xana coup (BOI21_xanadu)	C++17	100 / 100	94 ms	16496 KiB

xanadu

#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()

const int MAXN = 1e5 + 13;
const int INF = 1e9 + 7;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef array<array<int, 2>, 2> dparr;

int N;
vi edge[MAXN];
bitset<MAXN> flip;
dparr dp[MAXN]; //current state of vtx, whether vtx has been flipped.
int ans;

dparr comb(dparr l, dparr r)
{
    dparr res;
    FOR(i, 0, 2) FOR(j, 0, 2) res[i][j] = INF;
    FOR(i, 0, 2)
    {
        FOR(j, 0, 2)
        {
            FOR(k, 0, 2)
            {
                ckmin(res[k ^ j][i], r[i][j] + l[k][i]);
            }
        }
    }
    return res;
}
void dfs(int u, int p)
{
    for (int v : edge[u])
    {
        if (v == p) continue;
        dfs(v, u);
    }
    FOR(i, 0, 2)
    {
        FOR(j, 0, 2)
        {
            dp[u][i][j] = INF;
        }
    }
    dp[u][flip[u] ^ 1][1] = 1;
    dp[u][flip[u]][0] = 0;
    for (int v : edge[u])
    {
        if (v == p) continue;
        dp[u] = comb(dp[u], dp[v]);
    }
    return;
}

int32_t main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(4);
    cin >> N;
    FOR(i, 0, N - 1)
    {
        int u, v;
        cin >> u >> v; u--; v--;
        edge[u].PB(v);
        edge[v].PB(u);
    }
    FOR(i, 0, N)
    {
        bool b; cin >> b;
        flip[i] = b;
    }
    dfs(0, N);
    ans = min(dp[0][0][1], dp[0][0][0]);
    if (ans == INF)
    {
        cout << "impossible\n";
        return 0;
    }
    cout << ans << '\n';
    return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 15 / 15
• Subtask 3: 10 / 10
• Subtask 4: 40 / 40
• Subtask 5: 30 / 30