Codeforces-854C

链接:

  http://codeforces.com/contest/854/problem/C

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题目：

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.

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题意：

  有n架飞机，第i架飞机原本计划在第i分钟起飞，可是由于某种原因整个机场前k分钟是不能起飞的，每分钟只能起飞一架飞机，然后告诉你每架飞机每延误一分钟会造成的损失，问你如何安排飞机的起飞时间才能将损失降到最少（飞机不能提前起飞）。

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思路：

  维护一个最大化堆，这里采用优先队列。二级排序按照他们原本起飞时间的升序排列，然后先向队列中加入k+1个元素，然后每次pop出一个元素作为第一个起飞的，然后再push一个元素进去，直到队列为空，这样可以保证权值越大的飞机越优先起飞，就能保证最后损失最小。

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实现：

#include <cstdio>#include <queue>long long ans[int(1e7)+7];struct node {    long long t, w;    bool operator < (const node &tmp) const {        return w == tmp.w ? t > tmp.t : w < tmp.w;    }    node(long long t = 0, long long w = 0):t(t), w(w) {}}now;using namespace std;priority_queue<node> q;int main() {    long long n, k, tmp, result = 0;    scanf("%lld%lld",&n,&k);    for(int i=1 ; i<=n+k ; i++) {        if(i<=n) scanf("%lld", &tmp), q.emplace(i, tmp);        if(i>k) {            now = q.top(); q.pop();            result += now.w * (i - now.t);            ans[now.t] = i;        }    }    printf("%lld\n",result);    for(int i=1 ; i<n ; i++) printf("%lld ", ans[i]);    return 0*printf("%lld\n",ans[n]);}