Submission #903522

# Submission time Handle Problem Language Result Execution time Memory
903522 2024-01-11T08:19:24 Z GrindMachine Synchronization (JOI13_synchronization) C++17
100 / 100
1740 ms 30916 KB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
https://github.com/mostafa-saad/MyCompetitiveProgramming/blob/master/Olympiad/JOI/JOIOC-13-synchronization.txt

a[u] = #of distinct vals in node u
what happens when an edge is activated?

new_value = a[u]+a[v]-common (common = #of guys that are counted in both a[u] and a[v])
then set all values in the connected component of u to this value

how to find common?
common = a[u] when nodes (u,v) where connected the last time
this is just the value of a[u] when edge (u,v) was previously deleted

handle the following operations:
1) connect (u,v)
2) disconnect (u,v)
3) find the value of a[u]
4) set all values in the cc of u to x

maybe use link-cut tree to handle? (but too hard to implement)

easier way is to handle these ops using a fenwick tree
for each cc, rep = the node with the lowest depth

to find the rep of u, just b.s over the largest ances which we can reach by passing through active edges
finding the #of active edges on the path can be done with a fenwick tree (path sum queries)

maintain the value of a[u] just in the root, so the 3rd and 4th ops are done (just get/set the value at the root of the cc)
ops 1 and 2 can be done by point updates on the fenwick tree

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

template<typename T>
struct fenwick {
    int siz;
    vector<T> tree;

    fenwick() {

    }

    fenwick(int n) {
        siz = n;
        tree = vector<T>(n + 1);
    }

    int lsb(int x) {
        return x & -x;
    }

    void build(vector<T> &a, int n) {
        for (int i = 1; i <= n; ++i) {
            int par = i + lsb(i);
            tree[i] += a[i];

            if (par <= siz) {
                tree[par] += tree[i];
            }
        }
    }

    void pupd(int i, T v) {
        while (i <= siz) {
            tree[i] += v;
            i += lsb(i);
        }
    }

    T sum(int i) {
        T res = 0;

        while (i) {
            res += tree[i];
            i -= lsb(i);
        }

        return res;
    }

    T query(int l, int r) {
        if (l > r) return 0;
        T res = sum(r) - sum(l - 1);
        return res;
    }
};

vector<pll> adj[N];

struct lca_algo {
    // LCA template (for graphs with 1-based indexing)
 
    int LOG = 1;
    vector<int> depth;
    vector<vector<int>> up;
    vector<int> tin, tout;
    vector<int> edge_node;
    int timer = 1;
 
    lca_algo() {
 
    }
 
    lca_algo(int n) {
        lca_init(n);
    }
 
    void lca_init(int n) {
        while ((1 << LOG) < n) LOG++;
        up = vector<vector<int>>(n + 1, vector<int>(LOG, 1));
        depth = vector<int>(n + 1);
        tin = vector<int>(n + 1);
        tout = vector<int>(n + 1);
        edge_node = vector<int>(n + 1);
        lca_dfs(1, -1);
    }
 
    void lca_dfs(int node, int par) {
        tin[node] = timer++;
 
        for(auto [child,id] : adj[node]) {
            if (child == par) conts;
 
            up[child][0] = node;
            rep1(j, LOG - 1) {
                up[child][j] = up[up[child][j - 1]][j - 1];
            }
 
            depth[child] = depth[node] + 1;
            edge_node[id] = child;

            lca_dfs(child, node);
        }
 
        tout[node] = timer-1;
    }
 
    int lift(int u, int k) {
        rep(j, LOG) {
            if (k & (1 << j)) {
                u = up[u][j];
            }
        }
 
        return u;
    }
 
    int query(int u, int v) {
        if (depth[u] < depth[v]) swap(u, v);
        int k = depth[u] - depth[v];
        u = lift(u, k);
 
        if (u == v) return u;
 
        rev(j, LOG - 1, 0) {
            if (up[u][j] != up[v][j]) {
                u = up[u][j];
                v = up[v][j];
            }
        }
 
        u = up[u][0];
        return u;
    }
 
    int get_dis(int u, int v) {
        int lca = query(u, v);
        return depth[u] + depth[v] - 2 * depth[lca];
    }
 
    bool is_ances(int u, int v){
        return tin[u] <= tin[v] and tout[u] >= tout[v];
    }
};

void solve(int test_case)
{
    ll n,m,q; cin >> n >> m >> q;
    vector<pll> edges(n+5);

    rep1(i,n-1){
        ll u,v; cin >> u >> v;
        adj[u].pb({v,i}), adj[v].pb({u,i});
        edges[i] = {u,v};
    }

    vector<ll> a(n+5);
    rep1(i,n) a[i] = 1;

    lca_algo LCA(n);
    fenwick<ll> fenw(n+5);

    auto edge_change = [&](ll id, ll val){
        ll u = LCA.edge_node[id];
        fenw.pupd(LCA.tin[u],val);
        fenw.pupd(LCA.tout[u]+1,-val);
    };

    auto path_sum = [&](ll u, ll v){
        assert(LCA.is_ances(u,v));
        return fenw.sum(LCA.tin[v])-fenw.sum(LCA.tin[u]);
    };

    auto find_cc = [&](ll u){
        ll l = 0, r = LCA.depth[u];
        ll mx_ances = -1;

        while(l <= r){
            ll mid = (l+r) >> 1;
            ll ances = LCA.lift(u,mid);
            ll len = LCA.depth[u]-LCA.depth[ances];
            ll sum = path_sum(ances,u);

            if(sum == len){
                mx_ances = ances;
                l = mid+1;
            } 
            else{
                r = mid-1;
            }
        }

        assert(mx_ances != -1);
        return mx_ances;
    };

    vector<bool> active(n+5);
    vector<ll> edge_sub(n+5);

    while(m--){
        ll id; cin >> id;
        auto [u,v] = edges[id];
        ll pu = find_cc(u), pv = find_cc(v);

        if(!active[id]){
            active[id] = 1;
            ll new_val = a[pu]+a[pv]-edge_sub[id];
            edge_sub[id] = 0;
            edge_change(id,1);
            pu = find_cc(u);
            a[pu] = new_val;
        }
        else{
            active[id] = 0;
            edge_sub[id] = a[pu];
            edge_change(id,-1);
            ll pu_new = find_cc(u), pv_new = find_cc(v);
            a[pu_new] = a[pv_new] = a[pu];
        }
    }

    while(q--){
        ll u; cin >> u;
        ll pu = find_cc(u);
        ll ans = a[pu];
        cout << ans << endl;
    }
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}
# Verdict Execution time Memory Grader output
1 Correct 1 ms 2652 KB Output is correct
2 Correct 1 ms 2652 KB Output is correct
3 Correct 1 ms 2648 KB Output is correct
4 Correct 1 ms 2652 KB Output is correct
5 Correct 1 ms 2652 KB Output is correct
6 Correct 2 ms 2916 KB Output is correct
7 Correct 23 ms 4696 KB Output is correct
8 Correct 19 ms 4712 KB Output is correct
9 Correct 32 ms 4712 KB Output is correct
10 Correct 517 ms 26072 KB Output is correct
11 Correct 473 ms 25924 KB Output is correct
12 Correct 1624 ms 30256 KB Output is correct
13 Correct 186 ms 24980 KB Output is correct
14 Correct 260 ms 25344 KB Output is correct
# Verdict Execution time Memory Grader output
1 Correct 414 ms 25932 KB Output is correct
2 Correct 459 ms 27144 KB Output is correct
3 Correct 386 ms 29264 KB Output is correct
4 Correct 376 ms 29420 KB Output is correct
# Verdict Execution time Memory Grader output
1 Correct 1 ms 2904 KB Output is correct
2 Correct 1 ms 2904 KB Output is correct
3 Correct 1 ms 2652 KB Output is correct
4 Correct 1 ms 2652 KB Output is correct
5 Correct 1 ms 2904 KB Output is correct
6 Correct 4 ms 2908 KB Output is correct
7 Correct 61 ms 5124 KB Output is correct
8 Correct 1732 ms 30916 KB Output is correct
9 Correct 1643 ms 28456 KB Output is correct
# Verdict Execution time Memory Grader output
1 Correct 1740 ms 28216 KB Output is correct
2 Correct 523 ms 28084 KB Output is correct
3 Correct 522 ms 28436 KB Output is correct
4 Correct 563 ms 28464 KB Output is correct
# Verdict Execution time Memory Grader output
1 Correct 1 ms 2652 KB Output is correct
2 Correct 1 ms 2652 KB Output is correct
3 Correct 1 ms 2652 KB Output is correct
4 Correct 1 ms 2652 KB Output is correct
5 Correct 3 ms 2920 KB Output is correct
6 Correct 23 ms 4956 KB Output is correct
7 Correct 481 ms 26668 KB Output is correct
8 Correct 1702 ms 30624 KB Output is correct
9 Correct 210 ms 26016 KB Output is correct
10 Correct 322 ms 26460 KB Output is correct
11 Correct 507 ms 28628 KB Output is correct
12 Correct 504 ms 28492 KB Output is correct
13 Correct 527 ms 30616 KB Output is correct