Codeforces Round #433 （2）C:Planning

 C:   Planning

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 c_i 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, ..., c_n (1 ≤ c_i ≤ 10⁷), here c_i 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, ..., t_n (k + 1 ≤ t_i ≤ k + n), here t_i is the minute when the i-th flight must depart. If there are several optimal schedules, print any of them.

  题意： 看样例吧，不好描述。

  思路： 暴力写了一发，果断超时，改用个优先队列存数据就行。复杂度就是o（n+k） 了。

 

#include<bits/stdc++.h>
#define maxn 300100
using namespace std;


typedef long long ll;
typedef pair<int,int> pa;

int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        int vec[1000000];
        priority_queue<pa> que;
        int x;
        ll ans=0;
        for(int i=1;i<=n+k;i++)
        {
            if(i<=n)
            {
                scanf("%d",&x);
                que.push(pa(x,i));
            }
            if(i>k)
            {
                vec[que.top().second]=i;
                ans+=1ll*(i-que.top().second)*que.top().first;
                que.pop();
            }
        }
        printf("%lld\n",ans);
        for(int i=1;i<=n;i++)
            printf("%d%c",vec[i],i==n?'\n':' ');
    }
    return 0;
}