Submission #840280

#	Time	Username	Problem	Language	Result	Execution time	Memory
840280		mohammad_kilani	Closing Time (IOI23_closing)	C++17	100 / 100	221 ms	49376 KiB

closing

#include "closing.h"
#include <vector>
#include <bits/stdc++.h>
using namespace std;

int n , x , y;

long long k;
const int N = 200010;

vector< pair< int , int > > g[N];

long long a[N] , b[N];

void DFS(int node,int prnt,long long *a,long long dist){
    a[node] = dist;
    for(int i = 0 ;i < (int)g[node].size();i++){
        if(g[node][i].first != prnt)
            DFS(g[node][i].first , node , a , dist + g[node][i].second);
    }
}

vector< bool > onpath;

bool DFS(int node,int prnt){
    onpath[node] = true;
    if(node == y)
    return true;
    for(int i = 0 ;i < (int)g[node].size();i++){
        if(g[node][i].first == prnt) continue;
        if(DFS(g[node][i].first , node))
        return true;
    }
    onpath[node] = false;
    return false;
}

long long seg[4 * N];

int seg2[4 * N];

void update(int s,int e,int idx,int Idx,long long val,int val2){
    if(s > Idx || e < Idx)
        return;
    if(s == e){
       seg[idx] = val;
        seg2[idx] = val2;
        return;
    }
    update(s, ((s+e) >> 1) , (idx << 1) , Idx , val , val2);
    update(((s+e) >> 1) + 1,e,(idx << 1) + 1 , Idx , val , val2);
    seg[idx] = seg[(idx << 1)] + seg[(idx << 1) + 1];
    seg2[idx] = seg2[(idx << 1)] + seg2[(idx << 1) + 1];
}

int get(int s,int e,int idx,long long can){
    if(can >= seg[idx])
        return seg2[idx];
    if(s == e)
        return 0;
    if(can >= seg[(idx << 1)])
        return get(((s+e) >> 1) + 1,e,(idx << 1) + 1, can - seg[(idx << 1)]) + seg2[(idx << 1)];
    return get(s, ((s+e) >> 1),(idx << 1), can);
}

int max_score(int N, int X, int Y, long long K,std::vector<int> U, std::vector<int> V, std::vector<int> W)
{   
    n = N, x = X , y = Y , k = K;
    for(int i = 0;i < n;i++) g[i].clear();
    onpath = vector< bool > (n , 0);
    for(int i = 0 ;i < 4 * n;i++) seg[i] = 0, seg2[i] = 0;
    for(int i = 0 ;i < (int)U.size();i++){
        g[U[i]].push_back(make_pair(V[i] , W[i]));
        g[V[i]].push_back(make_pair(U[i] , W[i]));
    }
    DFS(X , -1 , a , 0);
    DFS(Y, - 1, b , 0);
    vector< long long > v;
    for(int i = 0 ;i < n;i++){
        if(a[i] > b[i]) swap(a[i] , b[i]);
        v.push_back(a[i]);
    }
    sort(v.begin(),v.end());
    int ans = 0;
    long long cur = 0;
    while(ans < (int)v.size() && v[ans] + cur <= k) cur += v[ans++];

    cur = 0;
    DFS(X , -1);

    vector< pair< long long , long long > > v2;
    vector< pair< long long , pair< long long , int > > > v3;

    priority_queue < long long > q1;

    int has = 0;
    for(int i = 0 ;i < n;i++){
        if(onpath[i]){
            cur += a[i];
            q1.push(a[i] - b[i]);
            has++;
        }
        else{
            //cout << i << " " << a[i] << " " << b[i] << " ready for two with " << b[i] << endl;
            v2.push_back(make_pair(b[i] , a[i]));
        }
    }
    if(cur > k) return ans;
    sort(v2.begin(),v2.end());
    for(int i = 0 ;i < (int)v2.size();i++){
        v3.push_back(make_pair(v2[i].second , make_pair(v2[i].first , i)));
    }
    sort(v3.begin(),v3.end());
    reverse(v3.begin(),v3.end());

    for(int i = 0 ;i < (int)v3.size();i++){
        update(0 , n - 1 , 1, v3[i].second.second , v3[i].second.first , 1);
    }

    ans = max(ans , has + get(0 , n - 1 , 1 ,k - cur) * 2);

    //cout << has << " " << get(0 , n - 1 , 1 ,k - cur) << " " << cur << endl;

    while((int)v3.size() > 0 || (int)q1.size() > 0){
        if(cur > k) break;
        if((int)v3.size() == 0 || (((int)q1.size() > 0) && (-q1.top() < v3.back().first))){
            cur -= q1.top();
            q1.pop();
            has++;
        }
        else{
            cur += v3.back().first;
            has++;
            q1.push(v3.back().first - v3.back().second.first);
            update(0 , n - 1 , 1 , v3.back().second.second , 0 , 0);
            v3.pop_back();
        }
        if(cur > k) break;
       // cout << has << " " << get(0 , n - 1 , 1 ,k - cur) << endl;
        ans = max(ans , has + get(0 , n - 1 , 1 ,k - cur) * 2);
    }
    return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 8 / 8
• Subtask 2: 9 / 9
• Subtask 3: 12 / 12
• Subtask 4: 14 / 14
• Subtask 5: 9 / 9
• Subtask 6: 11 / 11
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 17 / 17