Submission #916608

#	Time	Username	Problem	Language	Result	Execution time	Memory
916608		GrindMachine	Unique Cities (JOI19_ho_t5)	C++17	64 / 100	1016 ms	57888 KiB

joi2019_ho_t5

#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:
edi
https://codeforces.com/blog/entry/65042?#comment-491880

*/

const int MOD = 1e9 + 7;
const int N = 2e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

template<typename T>
struct lazysegtree {
    /*=======================================================*/

    struct data {
        ll mn,cnt;
    };

    struct lazy {
        ll a;
    };

    data d_neutral = {inf2,1};
    lazy l_neutral = {0};

    void merge(data &curr, data &left, data &right) {
        if(left.mn < right.mn) curr = left;
        else if(left.mn > right.mn) curr = right;
        else curr = {left.mn,left.cnt+right.cnt};
    }

    void create(int x, int lx, int rx, T v) {

    }

    void modify(int x, int lx, int rx, T v) {
        lz[x].a = v;
    }

    void propagate(int x, int lx, int rx) {
        ll v = lz[x].a;
        if(!v) return;

        tr[x].mn += v;

        if(rx-lx > 1){
            lz[2*x+1].a += v;
            lz[2*x+2].a += v;
        }

        lz[x] = l_neutral;
    }

    /*=======================================================*/

    int siz = 1;
    vector<data> tr;
    vector<lazy> lz;

    lazysegtree() {

    }

    lazysegtree(int n) {
        while (siz < n) siz *= 2;
        tr.assign(2 * siz, d_neutral);
        lz.assign(2 * siz, l_neutral);
    }

    void build(vector<T> &a, int n, int x, int lx, int rx) {
        if (rx - lx == 1) {
            if (lx < n) {
                create(x, lx, rx, a[lx]);
            }

            return;
        }

        int mid = (lx + rx) / 2;

        build(a, n, 2 * x + 1, lx, mid);
        build(a, n, 2 * x + 2, mid, rx);

        merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
    }

    void build(vector<T> &a, int n) {
        build(a, n, 0, 0, siz);
    }

    void rupd(int l, int r, T v, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return;
        if (lx >= l and rx <= r) {
            modify(x, lx, rx, v);
            propagate(x, lx, rx);
            return;
        }

        int mid = (lx + rx) / 2;

        rupd(l, r, v, 2 * x + 1, lx, mid);
        rupd(l, r, v, 2 * x + 2, mid, rx);

        merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
    }

    void rupd(int l, int r, T v) {
        rupd(l, r + 1, v, 0, 0, siz);
    }

    data query(int l, int r, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return d_neutral;
        if (lx >= l and rx <= r) return tr[x];

        int mid = (lx + rx) / 2;

        data curr;
        data left = query(l, r, 2 * x + 1, lx, mid);
        data right = query(l, r, 2 * x + 2, mid, rx);

        merge(curr, left, right);
        return curr;
    }

    data query(int l, int r) {
        return query(l, r + 1, 0, 0, siz);
    }
};

vector<ll> adj[N];
vector<ll> a(N);
pll diam = {-1,-1};

void dfs1(ll u, ll p, ll d){
    pll px = {d,u};
    amax(diam,px);
    trav(v,adj[u]){
        if(v == p) conts;
        dfs1(v,u,d+1);
    }
}

ll dis[N][2];
ll deepest[N][2];

void dfs2(ll u, ll p, ll t){
    trav(v,adj[u]){
        if(v == p) conts;
        dis[v][t] = dis[u][t]+1;
        dfs2(v,u,t);
        amax(deepest[u][t],deepest[v][t]+1);
    }
}

lazysegtree<ll> st(N);
vector<ll> ans(N);
vector<ll> mx_without(N);

void dfs3(ll u, ll p, ll d, ll t){
    ll mx = deepest[u][t];
    st.rupd(max(d-mx,0ll),d-1,1);
    auto [mn,mn_cnt] = st.query(0,d);
    assert(mn == 0);

    if(dis[u][t] >= dis[u][t^1]){
        ans[u] = mn_cnt-1;
    }

    st.rupd(max(d-mx,0ll),d-1,-1);

    mx = 0;

    trav(v,adj[u]){
        if(v == p) conts;
        mx_without[v] = mx;
        amax(mx,deepest[v][t]+1);
    }

    mx = 0;
    reverse(all(adj[u]));

    trav(v,adj[u]){
        if(v == p) conts;
        amax(mx_without[v],mx);
        amax(mx,deepest[v][t]+1);
    }

    reverse(all(adj[u]));

    trav(v,adj[u]){
        if(v == p) conts;
        mx = mx_without[v];
        st.rupd(d-mx,d-1,1);
        dfs3(v,u,d+1,t);
        st.rupd(d-mx,d-1,-1);
    }
}

void solve(int test_case)
{
    ll n,m; cin >> n >> m;
    rep1(i,n-1){
        ll u,v; cin >> u >> v;
        adj[u].pb(v), adj[v].pb(u);
    }
    rep1(i,n) cin >> a[i];

    dfs1(1,-1,0);
    ll s = diam.ss;
    diam = {-1,-1};
    dfs1(s,-1,0);
    ll t = diam.ss;

    dfs2(s,-1,0);
    dfs2(t,-1,1);

    st.rupd(0,n,-inf2);

    dfs3(s,-1,0,0);
    dfs3(t,-1,0,1);

    rep1(i,n){
        amin(ans[i],m);
        cout << ans[i] << endl;
    }
}

int main()
{
    fastio;

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

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

    return 0;
}

• Subtask 1: 0 / 4
• Subtask 2: 32 / 32
• Subtask 3: 32 / 32
• Subtask 4: 0 / 32