SDNU_ACM_ICPC_2020_Winter_Practice_1st C

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 = 20burles.

大意：有n个航班，分别在1~n起飞。现知道前k分钟不能起飞，那么有的飞机就要延误，现给出每个航班延误一分钟耗费的成本，重新安排飞机起飞，使成本最小。

思路

注意条件，所有飞机起飞时间不得早于之前的时间！！！
也就是在第k+i分钟起飞的飞机，只能在原定时间小于等于k+i的飞机中选择。并且每次选择其中延误成本最大的，就能把损失降到最低。

AC代码

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
struct pp
 {
     int yuan;
     int zhi;
      bool operator <(const pp& x)const{
         return zhi < x.zhi;///zhi大的出队
     }
 }a[300006],f;
priority_queue<pp> q;
int main()
{
    long long n,k,sum=0;
    cin>>n>>k;
    for(long long i=1;i<=n;i++)
    {
        scanf("%lld",&a[i].zhi);
        a[i].yuan=i;
    }
    for(long long i=1;i<=k;i++)
    {
        q.push(a[i]);///第1~k分钟的航班均可选择
    }
    for(long long i=k+1;i<=n+k;i++)///起飞时间
    {
        if(i<=n)
        {
            q.push(a[i]);
        }
        f=q.top();///第一个元素
        q.pop();///删除第一个元素
        sum=sum+(i-f.yuan)*f.zhi;
        a[f.yuan].yuan=i;
    }
    cout<<sum<<endl;
    for(long long i=1;i<=n;i++)
    {
       printf("%d ",a[i].yuan);
    }
    return 0;
}