C. Planning

传送门

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.

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.

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.

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.

题意：有n架飞机，他们原本的起飞时间依次是1~n，但前k分钟不能起飞，已知第i架飞机延误每1min花费为ci，且不能比原定时间早起飞问每架飞机在第几分钟起飞花费最少。

题解：因为不能比原定时间早起飞，我们先将在第k+i秒可以起飞的飞机放到优先队列里面，再将每分钟花费高的或者花费相同时间比较前的弹出来给他先安排时间起飞。（用set也可（set也是用于排序））

代码：

#include<bits/stdc++.h>

#define ll long long

using namespace std;

const int N = 3e5 + ;

struct node{

    int id,c;

    bool operator < (const node &x)const {

        if (c == x.c) return id < x.id;

        return c < x.c;

    }

}a[N],x;

int ans[N];

int main(){

    int n,k;

    scanf("%d%d",&n,&k);

    for (int i = ; i <= n; i++) {

        scanf("%d",&a[i].c);

        a[i].id = i;

    }

    ll sum = ;

    priority_queue<node> q;

    for (int i = ; i <= k; i++) q.push(a[i]);

    for (int i = k+; i <= k+n; i++) {

        if (i<=n) q.push(a[i]);

        x = q.top();

        q.pop();

        sum += 1ll*(i-x.id)*x.c;

        ans[x.id] = i;

    }

    printf("%lld\n", sum);

    for (int i = ; i <= n; i++)

        printf("%d%c", ans[i],i==n?'\n':' ');

    return ;

}

#include<bits/stdc++.h>

#define ll long long

using namespace std;

const int N = 3e5 + ;

struct node{

    int id,c;

}a[N],x;

bool cmp(node i,node j) {

    if (i.c == j.c) return i.id > j.id;

    return i.c>j.c;

}

int ans[N];

set<int> s;

int main(){

    int n,k;

    scanf("%d%d",&n,&k);

    for (int i = ; i <= n; i++) {

        scanf("%d",&a[i].c);

        a[i].id = i;

        s.insert(i+k);

    }

    sort(a+,a+n+,cmp);

    ll sum = ;

    for (int i = ; i <= n; i++) {

        int x = *s.lower_bound(a[i].id);

        sum += 1ll*(x-a[i].id)*a[i].c;

        ans[a[i].id] = x;

        s.erase(x);

    }

    printf("%lld\n", sum);

    for (int i = ; i <= n; i++)

        printf("%d%c", ans[i],i==n?'\n':' ');

    return ;

}