CodeForces - 853A Planning (优先队列,贪心)

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.

 

 新号第一次打cf,由于STL不熟练,手写优先队列写挂了,止步c题,还需努力.

 

 题意:本来安排了n架飞机,每架飞机有mi的重要度,第i架飞机的起飞时间原定为i,而现在在k时之前不能有任何飞机起飞,每个时间点只有1架飞机能起飞,损失被定义为(飞机现起飞时间-飞机原起飞时间)*该飞机的重要度.现在要求你排一张时间表,要求每架飞机都只能在原起飞时间及以后起飞,且使损失最小.

 

题解:显然是一道贪心,值越大的越先起飞,造成损失越小,因为假设a>b , a*n+b*(n+1)=(a+b)*n+b  (1) , a*(n+1)+b*n=(a+b)*n+a  (2).显然(1)>(2).所以枚举时间,将1-k之间的飞机先压入优先队列,接着k-n+k一边加进原定当时起飞的飞机,一边找到优先队列中的最大值即队顶元素,将它放在该位置,更新它的新的起飞时间.这样可以保证i时进来的飞机一定在i之后起飞.

 

代码如下:

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define N 300010
using namespace std;

struct NODE
{
    int i,w;
    bool operator <(const NODE&a)const
    {
        return w<a.w;
    }
}p[N];

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&p[i].w);
        p[i].i=i;
    }
    priority_queue<NODE>q;
    for(int i=1;i<=k;i++)
    {
        q.push(p[i]);
    }
    long long sum=0;
    for(int i=k+1;i<=n+k;i++)
    {
        if(i<=n)
        {
            q.push(p[i]);
        }
        NODE now=q.top();
        q.pop();
        sum+=(long long) (i-now.i)*now.w;
        p[now.i].i=i;
    }
    printf("%lld\n",sum);
    for(int i=1;i<=n;i++)
    {
        printf("%d ",p[i].i);
    }
    return 0;
}

 

 

 

每天刷题,身体棒棒!