Submission #840249

#	Time	Username	Problem	Language	Result	Execution time	Memory
840249		mohammad_kilani	Closing Time (IOI23_closing)	C++17	83 / 100	1068 ms	37180 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;
}

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 < (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;
    int ans2 = 0;

    DFS(X , -1);

    vector< long long > v1 , v3 , v4;
    vector< pair< long long , long long > > v2;

    int has = 0;
    for(int i = 0 ;i < n;i++){
        if(onpath[i]){
            cur += a[i];
            v1.push_back(b[i] - a[i]);
            has++;
        }
        else{
            v2.push_back(make_pair(a[i] , b[i]));
        }
    }

    ans2 = has;

    if(cur > k) return ans;

    sort(v1.begin(),v1.end());
    sort(v2.begin(),v2.end());

    for(int i = 0 ;i <= (int)v2.size();i++){
        v3.clear();
        v4 = v1;
        long long tmpcur = cur;
        for(int j = i ;j < (int)v2.size();j++){
            v3.push_back(v2[j].second);
        }
        for(int j = 0 ;j < i;j++){
            tmpcur += v2[j].first;
            v4.push_back(v2[j].second - v2[j].first);
        }
        if(tmpcur > k) break;
        sort(v3.begin(),v3.end());
        sort(v4.begin(),v4.end());

        long long tmptake = 0;
        int idx = 0;
        for(idx = 0; idx < (int)v4.size();idx++) tmptake += v4[idx];
        
        for(int j = 0 ;j <= (int)v3.size();j++){
            while(tmptake + tmpcur > k && idx > 0){
                tmptake -= v4[--idx];
            }
            if(tmpcur + tmptake > k) break;
            ans2 = max(ans2 , has + i +  idx + j * 2);
            if(j < (int)v3.size())
                tmpcur += v3[j];
        }
    }
    return max(ans , ans2);
}

• 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: 0 / 17