找出通过出售电影票可以收集的最大金额

给定一个整数N和一个数组座位 [] ，其中N是排队购买电影票的人数，座位 [i]是电影院^第i排的空座位数。任务是找到剧院老板通过向N人出售电影票可以赚取的最大金额。一张票的价格等于所有行中空座位的最大数量。
例子：

Input: seats[] = {1, 2, 4}, N = 3
Output: 9

Person	Empty Seats	Ticket Cost
1	1 2 4	4
2	1 2 3	3
3	1 2 2	2

4 + 3 + 2 = 9
Input: seats[] = {2, 3, 5, 3}, N = 4
Output: 15

编程需要懂一点英语

方法：这个问题可以通过使用优先队列来解决，该队列将存储每行空座位的数量，其中最大的将在顶部可用。

创建一个空的priority_queue q并遍历seats[]数组并将所有元素插入priority_queue。
初始化两个整数变量ticketSold = 0和ans = 0 ，它们将存储到目前为止售出的门票数量和总收藏金额。
现在检查 while ticketSold < N and q.top() > 0然后从 priority_queue 中删除顶部元素并通过添加优先级队列的顶部元素来更新 ans。还将这个最高值存储在变量temp 中，并将temp – 1插入到 priority_queue 中。
重复这些步骤，直到所有的人都卖完票并打印最终结果。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the maximum amount
// that can be collected by selling tickets
int maxAmount(int M, int N, int seats[])
{
 
    // Priority queue that stores
    // the count of empty seats
    priority_queue q;
 
    // Insert each array element
    // into the priority queue
    for (int i = 0; i < M; i++) {
        q.push(seats[i]);
    }
 
    // To store the  total
    // number of tickets sold
    int ticketSold = 0;
 
    // To store the total amount
    // of collection
    int ans = 0;
 
    // While tickets sold are less than N
    // and q.top > 0 then update the collected
    // amount with the top of the priority
    // queue
    while (ticketSold < N && q.top() > 0) {
        ans = ans + q.top();
        int temp = q.top();
        q.pop();
        q.push(temp - 1);
        ticketSold++;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int seats[] = { 1, 2, 4 };
    int M = sizeof(seats) / sizeof(int);
    int N = 3;
 
    cout << maxAmount(N, M, seats);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
    static int[] seats = new int[]{ 1, 2, 4 };
 
    // Function to return the maximum amount
    // that can be collected by selling tickets
    public static int maxAmount(int M, int N)
    {
 
        // Priority queue that stores
        // the count of empty seats
        PriorityQueue q =
            new PriorityQueue(Collections.reverseOrder());
     
        // Insert each array element
        // into the priority queue
        for (int i = 0; i < M; i++)
        {
            q.add(seats[i]);
        }
 
        // To store the total
        // number of tickets sold
        int ticketSold = 0;
     
        // To store the total amount
        // of collection
        int ans = 0;
     
        // While tickets sold are less than N
        // and q.top > 0 then update the collected
        // amount with the top of the priority
        // queue
        while (ticketSold < N && q.peek() > 0)
        {
            ans = ans + q.peek();
            int temp = q.peek();
            q.poll();
            q.add(temp - 1);
            ticketSold++;
        }
        return ans;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int M = seats.length;
        int N = 3;
     
        System.out.print(maxAmount(M, N));
    }
}
 
// This code is contributed by Sanjit_Prasad

Python 3

# Python 3 implementation of the approach
 
# Function to return the maximum amount
# that can be collected by selling tickets
def maxAmount(M, N, seats):
     
    # Priority queue that stores
    # the count of empty seats
    q = []
 
    # Insert each array element
    # into the priority queue
    for i in range(M):
        q.append(seats[i])
 
    # To store the total
    # number of tickets sold
    ticketSold = 0
 
    # To store the total amount
    # of collection
    ans = 0
 
    # While tickets sold are less than N
    # and q.top > 0 then update the collected
    # amount with the top of the priority
    # queue
    q.sort(reverse = True)
    while (ticketSold < N and q[0] > 0):
        ans = ans + q[0]
        temp = q[0]
        q = q[1:]
        q.append(temp - 1)
        q.sort(reverse = True)
        ticketSold += 1
 
    return ans
 
# Driver code
if __name__ == '__main__':
    seats = [1, 2, 4]
    M = len(seats)
    N = 3
 
    print(maxAmount(N, M, seats))
 
# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
     
    // Function to return the maximum amount
    // that can be collected by selling tickets
    static int maxAmount(int M, int N, int[] seats)
    {
       
        // Priority queue that stores
        // the count of empty seats
        List q = new List();
       
        // Insert each array element
        // into the priority queue
        for (int i = 0; i < M; i++) {
            q.Add(seats[i]);
        }
         
        q.Sort();
        q.Reverse();
       
        // To store the  total
        // number of tickets sold
        int ticketSold = 0;
       
        // To store the total amount
        // of collection
        int ans = 0;
       
        // While tickets sold are less than N
        // and q.top > 0 then update the collected
        // amount with the top of the priority
        // queue
        while (ticketSold < N && q[0] > 0) {
            ans = ans + q[0];
            int temp = q[0];
            q.RemoveAt(0);
            q.Add(temp - 1);
            q.Sort();
            q.Reverse();
            ticketSold++;
        }  
        return ans;
    }
 
  // Driver code
  static void Main()
  {
    int[] seats = { 1, 2, 4 };
    int M = seats.Length;
    int N = 3;
    Console.WriteLine(maxAmount(N, M, seats));
  }
}
 
// This code is contributed by divyesh072019.

Javascript

输出：