アイデア：小さなトップスタックを使用して、各マシンが空いたらすぐに、ジョブから最も長い作業時間のマシンを選択します。
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
class MechineNode
{
public:
MechineNode(int id, int start, int JobId = 0) {
this->ID = id;
this->start = start;
this->JobId = JobId;
};
MechineNode() {};
int ID, start, JobId;
};
class JobNode
{
public:
JobNode() {};
JobNode(int id, int time)
{
this->ID = id;
this->time = time;
}
int ID, time;
};
bool cmp1(JobNode& j1, JobNode& j2)
{
return j1.time <j2.time;
}
struct cmp
{
bool operator()(const MechineNode& m1, const MechineNode& m2)
{
return m1.start >m2.start;
}
};
void visit(vector<vector<int>>& v)
{
for (int i = 1; i <v.size(); i++)
{
cout << i << ": ";
for (int j = 1; j <v[i].size(); j++)
cout << v[i][j] << " ";
cout << endl;
}
}
void print(vector<JobNode>& v)
{
for (int i = 0; i < v.size(); i++)
cout << v[i].time << " ";
cout << endl;
}
int main()
{
int n, m;
while (cin >> n>>m)
{
priority_queue<MechineNode, vector<MechineNode>, cmp>p;
vector<JobNode>v;
JobNode j(-1,-1);
v.push_back(j);
//visit(rec);
vector<vector<int>> rec(m+1,vector<int>(1,0));
for (int i = 1; i <=n; i++)
{
int time;
cin >> time;
JobNode j(i,time);
v.push_back(j);
}
sort(v.begin(), v.end(),cmp1);
print(v);
for (int i = m; i >= 1; i--)
{
MechineNode t(i,0);
p.push(t);
}
for (int i=n; i>=1; i--)
{
MechineNode t;
t = p.top();
//cout << "t.id:" << t.ID << " t.start: " << t.start << endl;
int Mid = t.ID;
if (rec[Mid].size() == 1&&rec[Mid][0]!=0)
rec[Mid][0] = i;
else
rec[Mid].push_back(i);
p.pop();
t.start += v[i].time;
//cout << "t.id:" << t.ID << " t.start: " << t.start << endl;
t.JobId = i;
p.push(t);
}
visit(rec);
}
}
/*
7 4
5 8 7 6 10 4 2
*/