[백준, C++17] 1700 멀티탭 스케줄링, 근데 이제 for_each, distance, find를 곁들인

솔루션을 보고 풀었다.

 

https://jaimemin.tistory.com/759

백준 1700번 멀티탭 스케줄링

 

푸는데 for_each, distance, find 를 활용하고싶었다.

 

#include <iostream>
#include <algorithm>
#include <iterator>
#include <vector>
#include <array>

using namespace std;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    
    array<int, 100> schedule;
    array<int, 100> power; power.fill(0);

    int N, K, answer=0;
    cin >> N >> K;
    
    for_each(schedule.begin(), schedule.begin()+K, [](auto& x){ cin >> x; });
    
    for(int i=0; i<K; i++)
    {
        int device = schedule[i];
        bool ok = false;
        
        for(int j=0; j<N; j++)
        {
            if(power[j] == device)
            {
                ok = true;
                break;
            }
            if(power[j] == 0)
            {
                power[j] = device;
                ok = true;
                break;
            }
        }
        
        if(ok) continue;
        answer++;
        
        int max_next_schedule_distance = -1;
        int picked_device_position = -1;
        
        for(int j=0; j<N; j++)
        {
            int using_device = power[j];
            auto p = find(schedule.begin()+i+1, schedule.end(), using_device);

            //가장 마지막에 쓰이는 장치를 찾는다. 못찾으면 그걸 그냥 가장 마지막에 쓰인다고 봄
            int next_schedule_distance = distance(schedule.begin()+i+1, p);
            if(max_next_schedule_distance < next_schedule_distance)
            {
                max_next_schedule_distance = next_schedule_distance;
                picked_device_position = j;
            }
        }
        
        power[picked_device_position] = device;
    }
    cout << answer;
    return 0;
}