#433 - C. Planning（贪心，优先队列）

最新推荐文章于 2019-08-23 12:59:43 发布

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最新推荐文章于 2019-08-23 12:59:43 发布

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分类专栏： CF 文章标签：贪心优先队列

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本文解决了一个关于飞机因技术问题导致的延误调度问题。通过使用优先队列，实现了最小化总延误成本的目标。输入包括航班数量、延误时间和每分钟延误成本。

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题目连接：http://codeforces.com/contest/854/problem/C

题目描述

Description

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.

Sample Input

5 2
4 2 1 10 2

Sample Output

20
3 6 7 4 5

解题思路

题意：飞机要起飞，但是要延误K分钟，一共n架飞机
下面有n个数第i个数为第i分钟开始每秒损失n[i]金钱，例如4 2 1 10 2，第一分钟开始损失4金钱/分钟，第二分钟，第一架和第二架一共损失6金钱/分钟（都没起飞的情况下）
明白了题意就好办了，做一个优先队列，就是带排序的队列，int回爆，用long long

AC代码

#include<iostream>
#include<queue>
#define MAXN 1000010

using namespace std;

struct node {
    int id, cost;
//  node(int id, int cost):id(id), cost(cost) {};
    bool operator < (const node& n) const { //让队列排序 
        return cost < n.cost;
    }
}cnt[MAXN];

int main() {
    long long n, k; 
    cin >> n >> k;
    priority_queue<node> q;
    long long sum = 0;
    for(long long i = 1; i <= n; i++) {
        cin >> cnt[i].cost;
        cnt[i].id = i;
    } 
    for(long long i = 1; i <= k; i++) q.push(cnt[i]); //把延时的时候可以起飞的飞机放进队列 
    for(long long i = k+1; i <= n+k; i++) {
        if(i <= n) {
            q.push(cnt[i]);//要一个一个放进去，因为不能在本来起飞的时间前起飞 
        }
        node val = q.top();
        q.pop();
        sum += (long long) (i - val.id) * val.cost;//只要优先起飞花费最多的飞机一定是花费最少的 
        cnt[val.id].id = i;
    }
    cout << sum << endl;
    for(long long i = 1; i <= n; i++) {
        cout << cnt[i].id << " ";//这里没哟要求输出格式 
    }
    return 0;
}