제출 #635282

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
635282		Anthony_Liu	Job Scheduling (CEOI12_jobs)	C++11	100 / 100	390 ms	26360 KiB

jobs

#include<bits/stdc++.h>
using namespace std;
#define f first
#define s second
#define ll long long
#define pb push_back
#define pi pair <int,int>
#define vi vector <int>
#define size(x) (int)(x).size()
#define all(x) x.begin(), x.end()
void setIO(string name = "") {
	cin.tie(0)->sync_with_stdio(0);
	if (size(name)) {
		freopen((name + ".in").c_str(), "r", stdin);
		freopen((name + ".out").c_str(), "w", stdout);
	}
}

int N, D, M;

// test if it is possible to finish the jobs using given # of machines
// return: first: possible or not, second: if possible, the schedule for the jobs
pair<bool, vector<vi>> check(vector<pi> &jobs, int machineCount)
{
	vector<vi> schedule(N);
	int reqNum = 0;
	// we simulate from day 1 until the last day N
	// we move to the next day if all the machines are used or
	// there is no more job requests left on or before this day
	for (int day = 1; day <= N; day++)
	{
		for (int j = 0; j < machineCount; j++)
		{
			// if all jobs before and on this day are finished,
			// we can go to the next day, even if there are usable machines left
			// we can determine that since the vector jobs is sorted
			if (jobs[reqNum].f > day)
				break;
			// if the current date is before the deadline for the job
			// we can add this job to the schedule and move to the next job request
			if (jobs[reqNum].f + D >= day)
				schedule[day - 1].pb(jobs[reqNum++].s);
			// otherwise, it is not feasible due to deadline
			else
				return {false, schedule};

			// if we have processed all the requests, we have found a feasible sol
			if (reqNum == M)
				return {true, schedule};
		}
	}
	// if not all the requests can be processed within the given N days,
	// then it is not feasible
	return {false, schedule};
}

int main()
{
	cin.tie(0)->sync_with_stdio(false);

	cin >> N >> D >> M;
	vector<pi> jobs(M);
	for (int i = 0; i < M; i++)
	{
		int day;
		cin >> day;
		// first: request date, second: index [1..M]
		jobs[i] = {day, i+1};
	}
	// we sort the jobs by the request date in ascending order
	// sothat we can test them using isFeasible() in linear time whether they
	// can be done in given time using a certain amount of machines
	sort(all(jobs));

	vector<vi> result;
	// binary search on the number of machines for the minimum possible solution
	// left and right bound, l and r
	int l = 1, r = M;
	while (l < r)
	{
		int machineNum = (l + r) / 2;
		// test if the jobs would finish within the deadline
		// using the current # of machines, machineNum
		pair<bool, vector<vi>> curResult = check(jobs, machineNum);
		// if it's possible, we set the right bound as the tested machine number
		// and save the current schedule
		if (curResult.f)
		{
			r = machineNum;
			result = curResult.s;
		}
		// otherwise, we set the left bound to be the tested number + 1
		// and test the next machineNum again
		else
			l = machineNum + 1;
	}

	cout << l << "\n";
	for (int i = 0; i < N; i++)
	{
		for (int &idx : result[i])
			cout << idx << " ";
		cout << 0 << "\n";
	}
}

컴파일 시 표준 에러 (stderr) 메시지

jobs.cpp: In function 'void setIO(std::string)':
jobs.cpp:14:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   14 |   freopen((name + ".in").c_str(), "r", stdin);
      |   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
jobs.cpp:15:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   15 |   freopen((name + ".out").c_str(), "w", stdout);
      |   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 100 / 100