Submission #410947

#	Time	Username	Problem	Language	Result	Execution time	Memory
410947		JerryLiu06	Synchronization (JOI13_synchronization)	C++17	100 / 100	322 ms	24260 KiB

synchronization

#include <bits/stdc++.h>
 
using namespace std;
 
using ll = long long;
using db = long double;
using str = string;
 
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
 
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
 
#define mp make_pair
#define f first
#define s second

#define sz(x) (int)(x).size()
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define sor(x) sort(all(x))
#define ft front()
#define bk back()
#define pb push_back
#define pf push_front
 
#define lb lower_bound
#define ub upper_bound
 
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define R0F(i, a) ROF(i, 0, a)
#define EACH(a, x) for (auto& a : x)

const int MOD = 1e9 + 7;
const int MX = 1e5 + 10;
const ll INF = 1e18;

int N, M, Q, timer = 1; int U[MX], V[MX]; vi adj[MX]; bool active[MX];

int ans[MX], last[MX]; int in[MX], out[MX], lift[MX][20]; int BIT[MX];

// Preorder Traversal ~ O(N)

void DFS(int X, int P) {
    lift[X][0] = P; in[X] = timer++; ans[X] = 1;

    EACH(Y, adj[X]) if (Y != P) DFS(Y, X); out[X] = timer;
}
// Binary Indexed Tree

void update(int ind, int val) { while (ind <= timer) { BIT[ind] += val; ind += (ind & -ind); } }

int query(int ind) { int res = 0; while (ind > 0) { res += BIT[ind]; ind -= (ind & -ind); } return res; }

// Find the root of our connected component ~ O(lg^2 N)

int findRoot(int X) {
    R0F(i, 20) if (lift[X][i] && query(in[lift[X][i]]) == query(in[X])) X = lift[X][i]; return X;
}

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

    cin >> N >> M >> Q; F0R(i, N - 1) { cin >> U[i] >> V[i]; adj[U[i]].pb(V[i]), adj[V[i]].pb(U[i]); }

    DFS(1, 0); FOR(j, 1, 20) FOR(i, 1, N + 1) lift[i][j] = lift[lift[i][j - 1]][j - 1];
    
    FOR(i, 1, N + 1) update(in[i], -1), update(out[i], 1);
    
    F0R(i, M) { int ind; cin >> ind; ind--;
        int X = U[ind], Y = V[ind]; if (lift[X][0] == Y) swap(X, Y); // X is par of Y

        if (active[ind]) {
            ans[Y] = last[Y] = ans[findRoot(Y)]; update(in[Y], -1), update(out[Y], 1);
        }
        if (!active[ind]) {
            update(in[Y], 1), update(out[Y], -1); ans[findRoot(Y)] += ans[Y] - last[Y];
        }
        active[ind] = !active[ind];
    }
    F0R(i, Q) { int X; cin >> X; cout << ans[findRoot(X)] << "\n"; }
}

Compilation message (stderr)

synchronization.cpp: In function 'void DFS(int, int)':
synchronization.cpp:42:20: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
   42 | #define EACH(a, x) for (auto& a : x)
      |                    ^~~
synchronization.cpp:57:5: note: in expansion of macro 'EACH'
   57 |     EACH(Y, adj[X]) if (Y != P) DFS(Y, X); out[X] = timer;
      |     ^~~~
synchronization.cpp:57:44: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
   57 |     EACH(Y, adj[X]) if (Y != P) DFS(Y, X); out[X] = timer;
      |                                            ^~~
synchronization.cpp: In function 'int findRoot(int)':
synchronization.cpp:40:22: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
   40 | #define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
      |                      ^~~
synchronization.cpp:41:19: note: in expansion of macro 'ROF'
   41 | #define R0F(i, a) ROF(i, 0, a)
      |                   ^~~
synchronization.cpp:68:5: note: in expansion of macro 'R0F'
   68 |     R0F(i, 20) if (lift[X][i] && query(in[lift[X][i]]) == query(in[X])) X = lift[X][i]; return X;
      |     ^~~
synchronization.cpp:68:89: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
   68 |     R0F(i, 20) if (lift[X][i] && query(in[lift[X][i]]) == query(in[X])) X = lift[X][i]; return X;
      |                                                                                         ^~~~~~

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 10 / 10
• Subtask 4: 20 / 20
• Subtask 5: 40 / 40