Codeforces Round #433 (Div. 2, C. Planning （贪心+简单实现）

C. Planning
time limit per test
1 second
memory limit per test
512 megabytes
input
standard input
output
standard output

Helen works in Metropolis airport. She is responsible for creating a departure schedule. There are n flights that must depart today, the i-th of them is planned to depart at the i-th minute of the day.

Metropolis airport is the main transport hub of Metropolia, so it is difficult to keep the schedule intact. This is exactly the case today: because of technical issues, no flights were able to depart during the first k minutes of the day, so now the new departure schedule must be created.

All n scheduled flights must now depart at different minutes between (k + 1)-th and (k + n)-th, inclusive. However, it’s not mandatory for the flights to depart in the same order they were initially scheduled to do so — their order in the new schedule can be different. There is only one restriction: no flight is allowed to depart earlier than it was supposed to depart in the initial schedule.

Helen knows that each minute of delay of the i-th flight costs airport ci burles. Help her find the order for flights to depart in the new schedule that minimizes the total cost for the airport.
Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 300 000), here n is the number of flights, and k is the number of minutes in the beginning of the day that the flights did not depart.

The second line contains n integers c1, c2, …, cn (1 ≤ ci ≤ 107), here ci is the cost of delaying the i-th flight for one minute.
Output

The first line must contain the minimum possible total cost of delaying the flights.

The second line must contain n different integers t1, t2, …, tn (k + 1 ≤ ti ≤ k + n), here ti is the minute when the i-th flight must depart. If there are several optimal schedules, print any of them.
Example
input

5 2
4 2 1 10 2

output

20
3 6 7 4 5

Note

Let us consider sample test. If Helen just moves all flights 2 minutes later preserving the order, the total cost of delaying the flights would be (3 - 1)·4 + (4 - 2)·2 + (5 - 3)·1 + (6 - 4)·10 + (7 - 5)·2 = 38 burles.

However, the better schedule is shown in the sample answer, its cost is (3 - 1)·4 + (6 - 2)·2 + (7 - 3)·1 + (4 - 4)·10 + (5 - 5)·2 = 20 burles.

题意：给你n个飞机，k。按出发顺序1-n给你每个飞机拖延1分钟所要给的分数。让你重新排列飞机的飞行表。前1-k分钟不能起飞。求一个出发顺序，让总分数最少。

原本打算用set的，结果错了。。
看了别人是这样的，非常好。。

int ans[N];priority_queue<pii>q;int c[N];int main(){    int n,k;sf("%d%d",&n,&k);    rep(i,1,n){ sf("%d",&c[i]); }    int tp=1;    LL tot=0;    for(int t=k+1;t<=k+n;++t){        while(tp<=t&&tp<=n){            q.push(MP(c[tp],tp));            ++tp;        }        pii x=q.top();q.pop();        ans[x.second]=t;        tot+=1LL*x.first*(1LL*t-x.second);    }    cout<<tot<<'\n';    rep(i,1,n){        pf("%d%c",ans[i],i==n?'\n':' ');    }}