​​http://www.elijahqi.win/archives/714​​​
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

把在当前时间点之前的点全部扔入优先队列，维护大根堆，然后每次取出堆中最大的元素即可

因为假如这个元素已经过了他应该起飞的点，那么他顺延造成的损失将会更大

感谢zhx 在昨晚让我学习了pair 的用法，pair相当于是一个类型pair

#include<cstdio>
#include<queue>
#define pa std::pair<int,int>
#define N 330000
std::priority_queue<pa,std::vector<pa>,std::less<pa> > q;
int ans[N],c[N],n,k;
int main(){
    //freopen("cfc.in","r",stdin);
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;++i) scanf("%d",&c[i]);
    for (int i=1;i<=k;++i) q.push(std::make_pair(c[i],i)); 
    for (int i=k+1;i<=n+k;++i){
        if (i<=n) q.push(std::make_pair(c[i],i));
        ans[q.top().second]=i;q.pop();
    }
    long long ans1=0;
    for (int i=1;i<=n;++i) ans1+=(long long)c[i]*(ans[i]-i);
    printf("%lld\n",ans1);
    for (int i=1;i<=n;++i) printf("%d ",ans[i]);
    return 0;
}