Codeforces Round #433 (Div. 2) C. Planning（贪心）

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 = 20burles.

题解：

题意：

你很有可能赶不上飞机，你每拖延一分钟各种飞机要你交一定数额的拖延费，n为飞机数目，k为你要从第k秒开始赶飞机（当然只是单位时间，这里直接用秒了，一秒一辆真快），然后你可以任意选择登机顺序，使得你的拖延费最少，输入的是第i秒开始登机时每拖延一秒的费用，注意只能在登机开始后才能登机，不能提前登机

思路：

用一个优先队列，把可以登机的飞机加入套餐，由于这个费用每一秒增加是固定的，那么我们就选择每秒要交费用最多的那个，然后啪啪啪循环到底就行了

代码：

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
#include<algorithm>
using namespace std;
#define INF 100861111
#define eps 1e-7
#define lson k*2
#define rson k*2+1
#define ll long long
struct node
{
    int t;
    int v;
    friend bool operator<(node x,node y)
    {
        return x.v<y.v;
    }
}a[300005];
ll vis[300005];
priority_queue<node>q;
int main()
{
    int i,j,n,k,d,ans=0,temp;
    ll s=0;
    node now;
    scanf("%d%d",&n,&k);
    for(i=1;i<=n;i++)
    {
        vis[i]=0;
        scanf("%d",&a[i].v);
        a[i].t=i;
    }
    while(!q.empty())
        q.pop();
    d=min(n,k);
    d++;
    if(d<n)
        temp=d;
    else
        temp=n;
    for(i=1;i<=temp;i++)
    {
        q.push(a[i]);
    }
    for(i=1;i<=n;i++)
    {
        now=q.top();
        q.pop();
        vis[now.t]=d;
        s+=(ll)(d-now.t)*now.v;
        d++;
        if(temp<n)
        {
            q.push(a[temp+1]);
        }
        if(d<n)
            temp=d;
        else
            temp=n;

    }
    printf("%lld\n",s);
    printf("%lld",vis[1]);
    for(i=2;i<=n;i++)
        printf(" %lld",vis[i]);
    printf("\n");
    return 0;
}