Submission #1221777

#	Time	Username	Problem	Language	Result	Execution time	Memory
1221777		Theo830	Tree (IOI24_tree)	C++20	18 / 100	62 ms	22432 KiB

tree

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF = 1e9+7;
const ll MOD = 998244353;
typedef pair<ll,ll> ii;
#define iii pair<ll,ii>
#define id pair<ll,vector<ll> >
#define f(i,a,b) for(ll i = a;i < b;i++)
#define pb push_back
#define vll vector<ll>
#define F first
#define S second
#define all(x) (x).begin(), (x).end()
///I hope I will get uprating and don't make mistakes
///I will never stop programming
///sqrt(-1) Love C++
///Please don't hack me
///@TheofanisOrfanou Theo830
///Think different approaches (bs,dp,greedy,graphs,shortest paths,mst)
///Stay Calm
///Look for special cases
///Beware of overflow and array bounds
///Think the problem backwards
///Training
#include "tree.h"
int n;
long long leaf;
std::vector<int> p, w;
vector<vector<int> >adj;
vector<ll>ex;
ll pre[200005];
ll solve(ll idx,bool b){
    ll sum = w[idx] ^ 1;
    if(adj[idx].empty()){
        sum = 1;
    }
    if(w[idx] == 0){
        for(auto x:adj[idx]){
            solve(x,1);
        }
    }
    else{
        for(auto x:adj[idx]){
            sum += solve(x,0);
        }
    }
    b &= w[idx];
    if(b){
        ex.pb(sum);
    }
    return sum;
}
void init(std::vector<int> P, std::vector<int> W) {
    p = P;
    w = W;
    leaf = 0;
    ex.clear();
    n = (int)p.size();
    adj.assign(n+5,vector<int>());
    f(i,1,n){
        adj[p[i]].pb(i);
    }
    f(i,0,n){
        if(adj[i].size() == 0 && w[i]){
            leaf++;
        }
    }
    solve(0,1);
    sort(all(ex));
    pre[0] = 0;
    f(i,0,ex.size()){
        pre[i+1] = pre[i] + ex[i];
    }
}
long long query(int L, int R){
    ll ans = leaf * L;
    ll l = 0,r = ex.size()-1;
    ll pos = ex.size();
    while(l <= r){
        ll mid = (l+r)/2;
        if(ex[mid] * L > R){
            r = mid - 1;
            pos = min(pos,mid);
        }
        else{
            l = mid + 1;
        }
    }
    ans -= (ex.size() - pos) * R;
    ans += (pre[ex.size()] - pre[pos]) * L;
    return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 10
• Subtask 2: 0 / 13
• Subtask 3: 0 / 18
• Subtask 4: 7 / 7
• Subtask 5: 11 / 11
• Subtask 6: 0 / 22
• Subtask 7: 0 / 19