Submission #340186

#	Time	Username	Problem	Language	Result	Execution time	Memory
340186		blue	Kitchen (BOI19_kitchen)	C++11	100 / 100	41 ms	748 KiB

kitchen

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

//9 pts: 1 <= M <= 2
/*
int main()
{
    int N, M, K;
    cin >> N >> M >> K;

    int A[N], B[M];
    int Asum = 0;
    bool any1 = 0;
    for(int i = 0; i < N; i++)
    {
        cin >> A[i];
        Asum += A[i];
        if(A[i] == 1) any1 = 1;
    }
    for(int j = 0; j < M; j++) cin >> B[j];

    if(M == 1)
    {
        if(K == 1 && B[0] >= Asum) cout << B[0] - Asum << '\n';
        else cout << "Impossible\n";
    }
    else if(M == 2)
    {
        if(K == 1)
        {
            if(B[0] > B[1]) swap(B[0], B[1]);
            if(Asum <= B[0]) cout << B[0] - Asum << '\n';
            else if(Asum <= B[1]) cout << B[1] - Asum << '\n';
            else if(Asum <= B[0] + B[1]) cout << B[0] + B[1] - Asum << '\n';
            else cout << "Impossible\n";
        }
        else if(K == 2)
        {
            if(!any1 && B[0] >= N && B[1] >= N && B[0] + B[1] >= Asum) cout << B[0] + B[1] - Asum << '\n';
            else cout << "Impossible\n";
        }
        else cout << "Impossible\n";
    }
}
*/

//20 pts: K = 1
/*
int main()
{
    int N, M, K;
    cin >> N >> M >> K;

    int A[N], B[M];
    int Asum = 0;
    for(int i = 0; i < N; i++)
    {
        cin >> A[i];
        Asum += A[i];
    }
    for(int j = 0; j < M; j++) cin >> B[j];

    vector<int> possible(90001, 0);
    possible[0] = 1;
    int res = 90000;
    for(int j = 0; j < M; j++)
    {
        for(int i = 90000; i >= B[j]; i--)
        {
            if(!possible[i - B[j]]) continue;
            possible[i] = 1;
            if(i >= Asum) res = min(res, i - Asum);
        }
    }
    if(res == 90000) cout << "Impossible\n";
    else cout << res << '\n';
}
*/

/*
Condition for value K, meals 1, 2, ... N and chefs* 1, 2, ... y to be compatible  (* - chefs are re-numbered)
A[i] >= K
sum{j:chefs | max(B[j], N)} >= N*K
sum{j:chefs | B[j]} >= sum{i:meals | A[i]}
To minimize: sum{j:chefs | B[j]} - sum{i:meals | A[i]
*/

//100 pts
int main()
{
    int N, M, K;
    cin >> N >> M >> K;

    int A[N], B[M];
    bool impossible = 0;
    int Asum = 0;
    for(int i = 0; i < N; i++)
    {
        cin >> A[i];
        Asum += A[i];
        if(A[i] < K) impossible = 1;
    }
    for(int j = 0; j < M; j++) cin >> B[j];
    if(impossible)
    {
        cout << "Impossible\n";
        return 0;
    }
    //vector<bool> chef_sum(90001, 0);
    //chef_sum[0] = 1;
    vector<int> variety_sum(90001, -1); //sum over all dishes for number of different chefs for given total chef time sum
    variety_sum[0] = 0;

    for(int j = 0; j < M; j++)
    {
        for(int i = 90000; i >= B[j]; i--)
        {
            if(variety_sum[i - B[j]] >= 0)
            {
                variety_sum[i] = max(variety_sum[i], variety_sum[i - B[j]] + min(B[j], N));
            }
        }
    }

    for(int i = Asum; i <= 90000; i++)
    {
        if(variety_sum[i] >= N*K)
        {
            cout << i - Asum << '\n';
            return 0;
        }
    }
    cout << "Impossible\n";
}

• Subtask 1: 9 / 9
• Subtask 2: 22 / 22
• Subtask 3: 20 / 20
• Subtask 4: 21 / 21
• Subtask 5: 28 / 28