Submission #813031

#	Time	Username	Problem	Language	Result	Execution time	Memory
813031		stefanopulos	Vinjete (COI22_vinjete)	C++17	56 / 100	185 ms	20852 KiB

Main

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ldb;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ldb,ldb> pdd;

#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for(auto& a : x)
 
#define sz(a) (int)(a).size()
#define fi first
#define se second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

const int mod = 1000000007;
const int inf = 1e9 + 5;
const int mxN = 200005; 

int n, m;

vector<array<int,3>> g[mxN];

int mn[4 * mxN];
int cnt[4 * mxN];
int lenj[4 * mxN];
void build(int v, int tl, int tr){
    if(tl == tr){
        mn[v] = 0; cnt[v] = 1;
        return; 
    }
    int mid = (tl + tr) / 2;
    build(v * 2, tl, mid); build(v * 2 + 1, mid + 1, tr);
    mn[v] = min(mn[v * 2], mn[v * 2 + 1]);
    cnt[v] = (mn[v] == mn[v * 2] ? cnt[v * 2] : 0) 
    + (mn[v] == mn[v * 2 + 1] ? cnt[v * 2 + 1] : 0);
}

void propagate(int v, int tl, int tr){
    if(lenj[v] != 0){
        mn[v] += lenj[v];
        if(tl != tr){
            lenj[v * 2] += lenj[v];
            lenj[v * 2 + 1] += lenj[v];
        }
        lenj[v] = 0;
    }
}

void lazyupd(int v, int tl, int tr, int l, int r, int val){
    propagate(v, tl, tr);
    if(tl > tr || l > tr || tl > r)return;
    if(tl >= l && tr <= r){
        lenj[v] += val;
        propagate(v, tl, tr);
        return;
    }
    int mid = (tl + tr) / 2;
    lazyupd(v * 2, tl, mid, l, r, val); lazyupd(v * 2 + 1, mid + 1, tr, l, r, val);
    mn[v] = min(mn[v * 2], mn[v * 2 + 1]);
    cnt[v] = (mn[v] == mn[v * 2] ? cnt[v * 2] : 0) 
    + (mn[v] == mn[v * 2 + 1] ? cnt[v * 2 + 1] : 0);
}

int res[mxN];
void dfs(int v, int p){

    res[v] = m - (mn[1] == 0 ? cnt[1] : 0);
    for(auto c : g[v]){
        int u = c[0];
        int l = c[1];
        int r = c[2];
        if(u != p){
            lazyupd(1,1,m,l,r,1);
            dfs(u, v);
            lazyupd(1,1,m,l,r,-1);
        }
    }

}

int main(){
    cin.tie(0)->sync_with_stdio(0);

    cin >> n >> m;
    ff(i,1,n - 1){
        int u, v, l, r;
        cin >> u >> v >> l >> r;
        g[u].pb({v, l, r});
        g[v].pb({u, l, r});
    }

    build(1,1,m); dfs(1, -1);

    ff(i,2,n)cout << res[i] << '\n';

    return 0;
}
/*



// probati bojenje sahovski
*/
 
 
 
 
 

• Subtask 1: 11 / 11
• Subtask 2: 0 / 13
• Subtask 3: 16 / 16
• Subtask 4: 29 / 29
• Subtask 5: 0 / 31