Submission #374477

#	Time	Username	Problem	Language	Result	Execution time	Memory
374477		ne4eHbKa	Svjetlo (COCI20_svjetlo)	C++17	110 / 110	855 ms	124500 KiB

svjetlo

#include <bits/stdc++.h>
using namespace std;
#ifndef _LOCAL
//#pragma GCC optimize("O3,Ofast")
#else
#pragma GCC optimize("O0")
#endif
template<typename t> inline void umin(t &a, const t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, const t b) {a = max(a, b);}
typedef pair<int, int> pii;
typedef long long ll;
typedef long double ld;
typedef int8_t i8;
ll time() {return chrono::system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
#define ft first
#define sd second
#define len(f) int((f).size())
#define bnd(f) (f).begin(), (f).end()
#define _ <<' '<<
#define int ll
const int inf = 1e9 + 5;
const ll inf64 = 4e18 + 5;
const int md = 998244353;
namespace MD {
    void add(int &a, const int b) {if((a += b) >= md) a -= md;}
    void sub(int &a, const int b) {if((a -= b) < 0) a += md;}
    int prod(const int a, const int b) {return ll(a) * b % md;}
};

int n;
const int N = 5e5 + 5;
vector<int> g[N], h;
i8 p[N];
bool unlit[N], mk[N];
int s[2][N], t[2][N], f[2][N], e[N], T[2][2][N], S[2][3][N], rT[2][2][N];

int min(int a, int b, int c) {return min(a, min(b, c));}
int min(int a, int b, int c, int d) {return min(min(a, d), min(b, c));}

void solve() {
    cin >> n;
#ifdef KEK
    memset(mk, 0, N);
    for(int i = 0; i < n; ++i)
        g[i].clear();
#endif
    for(int i = 0; i < n; ++i) {
        char c; cin >> c;
        p[i] = c == '0';
        e[i] = 0;
    }
    for(int i = 1; i < n; ++i) {
        int u, v;
        cin >> u >> v;
        --u, --v;
        g[u].push_back(v);
        g[v].push_back(u);
        ++e[u], ++e[v];
    }
    queue<int> q;
    for(int i = 0; i < n; ++i)
        if(e[i] == 1) q.push(i);
    int cnt = 0;
    for(;;) {
        ++cnt;
        assert(!q.empty());
        int v = q.front();
        q.pop();
        mk[v] = true;
        unlit[v] = p[v];
        h.clear();
        for(int u : g[v]) {
            if(--e[u] == 1)
                q.push(u);
            if(mk[u] && unlit[u]) {
                unlit[v] = true;
                h.push_back(u);
            }
        }
        if(!unlit[v]) continue;
        int k = len(h);
        f[0][v] = +inf;
        f[1][v] = 1;
        for(int u : h) {
            int f0 = min(f[1][v] + f[0][u] + 1,
                         f[0][v] + f[1][u] + 3);
            int f1 = min(f[0][v] + f[0][u] + 1,
                         f[1][v] + f[1][u] + 3);
            f[0][v] = min(+inf, f0);
            f[1][v] = min(+inf, f1);
        }

        for(int i : {0, 1}) for(int j : {0, 1})
            fill(T[i][j], T[i][j] + k + 1, +inf);
        T[1][0][0] = 1;
        for(int i = 0; i < k; ++i) {
            for(int p : {0, 1}) {
                for(int z : {0, 1})
                    for(int ff : {0, 1})
                        umin(T[p ^ ff ^ 1][z][i + 1],
                             T[p][z][i] + f[ff][h[i]] + 1 + (ff << 1));
                for(int e : {0, 1})
                    umin(T[p ^ e][1][i + 1],
                         T[p][0][i] + t[e][h[i]] + (e << 1));
            }
        }

        t[0][v] = min(T[0][0][k], T[0][1][k], f[0][v]);
        t[1][v] = min(T[1][0][k], T[1][1][k], f[1][v]);
        for(int i : {0, 1}) for(int j : {0, 1, 2})
            fill(S[i][j], S[i][j] + k + 1, +inf);
        S[1][0][0] = 1;
        for(int i = 0; i < k; ++i) {
            for(int p : {0, 1}) {
                for(int z : {0, 1, 2})
                    for(int ff : {0, 1})
                        umin(S[p ^ ff ^ 1][z][i + 1],
                             S[p][z][i] + f[ff][h[i]] + 1 + (ff << 1));
                for(int z : {0, 1})
                    for(int e : {0, 1})
                        umin(S[p ^ e][z + 1][i + 1],
                            S[p][z][i] + t[e][h[i]] + (e << 1));
            }
        }
        s[0][v] = min(S[0][0][k], S[0][1][k], S[0][2][k], t[0][v]);
        s[1][v] = min(S[1][0][k], S[1][1][k], S[1][2][k], t[1][v]);

        for(int i : {0, 1}) for(int j : {0, 1})
            fill(rT[i][j], rT[i][j] + k + 1, +inf);
        rT[0][0][0] = 1;
        for(int i = 0; i < k; ++i) {
            for(int p : {0, 1}) {
                for(int z : {0, 1})
                    for(int ff : {0, 1})
                        umin(rT[p ^ ff ^ 1][z][i + 1],
                             rT[p][z][i] + f[ff][h[k - 1 - i]] + 1 + (ff << 1));
                for(int e : {0, 1})
                    umin(rT[p ^ e][1][i + 1],
                         rT[p][0][i] + t[e][h[k - 1 - i]] + (e << 1));
            }
        }

        for(int i = 0; i < k; ++i) {
            for(int fi : {0, 1}) {
                for(int se : {0, 1}) {
                    ll P = s[1][h[i]];
                    P += T[fi][0][i];
                    P += rT[se][0][k - 1 - i];
                    if(P > +inf) continue;
                    umin(s[fi ^ se][v], int(P));
                }
            }
        }
        if(p[v]) {
            swap(f[0][v], f[1][v]);
            swap(t[0][v], t[1][v]);
            swap(s[0][v], s[1][v]);
        }
        if(cnt == n) {
            cout << s[0][v] << endl;
            return;
        }
    }
}

signed main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#ifndef _LOCAL
//    freopen("file.in", "r", stdin);
//    freopen("file.out", "w", stdout);
#else
    system("color a");
    freopen("in.txt", "r", stdin);
    int t; cin >> t;
    while(t--)
#endif
    solve();
}

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