给定阵列的座位[]其中座椅[i]为的空闲座位的第i行中的用于板球比赛运动场数量。有N个人在排队等着买票。每个座位的成本等于它所属排的空座位数。任务是通过将门票出售给N个人来最大化利润。
例子：
Input: seats[] = {2, 1, 1}, N = 3
Output: 4
Person 1: Sell the seat in the row with
2 vacant seats, seats = {1, 1, 1}
Person 2: All the rows have 1 vacant
seat each, seats[] = {0, 1, 1}
Person 3: seats[] = {0, 0, 1}
Input: seats[] = {2, 3, 4, 5, 1}, N = 6
Output: 22
做法：为了利润最大化，票必须是空座位数最多的一排座位，当一个座位刚售完时，该排空座位数减1 .所有的人都可以卖一张座位票，直到有空位。这可以在 priority_queue 的帮助下有效计算。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the maximized profit
int maxProfit(int seats[], int k, int n)
{
// Push all the vacant seats
// in a priority queue
priority_queue pq;
for (int i = 0; i < k; i++)
pq.push(seats[i]);
// To store the maximized profit
int profit = 0;
// To count the people that
// have been sold a ticket
int c = 0;
while (c < n) {
// Get the maximimum number of
// vacant seats for any row
int top = pq.top();
// Remove it from the queue
pq.pop();
// If there are no vacant seats
if (top == 0)
break;
// Update the profit
profit = profit + top;
// Push the updated status of the
// vacant seats in the current row
pq.push(top - 1);
// Update the count of persons
c++;
}
return profit;
}
// Driver code
int main()
{
int seats[] = { 2, 3, 4, 5, 1 };
int k = sizeof(seats) / sizeof(int);
int n = 6;
cout << maxProfit(seats, k, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG {
// Function to return the maximized profit
static int maxProfit(int seats[], int k, int n)
{
// Push all the vacant seats
// in a priority queue
PriorityQueue pq;
pq = new PriorityQueue<>(Collections.reverseOrder());
for(int i = 0; i < k; i++)
pq.add(seats[i]);
// To store the maximized profit
int profit = 0;
// To count the people that
// have been sold a ticket
int c = 0;
while (c < n)
{
// Get the maximimum number of
// vacant seats for any row
int top = pq.remove();
// If there are no vacant seats
if (top == 0)
break;
// Update the profit
profit = profit + top;
// Push the updated status of the
// vacant seats in the current row
pq.add(top - 1);
// Update the count of persons
c++;
}
return profit;
}
// Driver Code
public static void main(String args[])
{
int seats[] = { 2, 3, 4, 5, 1 };
int k = seats.length;
int n = 6;
System.out.println(maxProfit(seats, k ,n));
}
}
// This code is contributed by rutvik_56 Python3
# Python3 implementation of the approach
# Function to return the maximized profit
def maxProfit(seats, k, n) :
# Push all the vacant seats
# in a priority queue
pq = [];
for i in range(k) :
pq.append(seats[i]);
# for maintaining the property of max heap
pq.sort(reverse = True);
# To store the maximized profit
profit = 0;
# To count the people that
# have been sold a ticket
c = 0;
while (c < n) :
# for maintaining the property of max heap
pq.sort(reverse = True);
# Get the maximimum number of
# vacant seats for any row
top = pq[0];
# Remove it from the queue
pq.pop(0);
# If there are no vacant seats
if (top == 0) :
break;
# Update the profit
profit = profit + top;
# Push the updated status of the
# vacant seats in the current row
pq.append(top - 1);
# Update the count of persons
c += 1;
return profit;
# Driver code
if __name__ == "__main__" :
seats = [ 2, 3, 4, 5, 1 ];
k = len(seats);
n = 6;
print(maxProfit(seats, k, n));
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
// Function to return the maximized profit
static int maxProfit(int[] seats, int k, int n)
{
// Push all the vacant seats
// in a priority queue
List pq = new List();
for(int i = 0; i < k; i++)
pq.Add(seats[i]);
// To store the maximized profit
int profit = 0;
// To count the people that
// have been sold a ticket
int c = 0;
while (c < n)
{
// Get the maximimum number of
// vacant seats for any row
pq.Sort();
pq.Reverse();
int top = pq[0];
// Remove it from the queue
pq.RemoveAt(0);
// If there are no vacant seats
if (top == 0)
break;
// Update the profit
profit = profit + top;
// Push the updated status of the
// vacant seats in the current row
pq.Add(top - 1);
// Update the count of persons
c++;
}
return profit;
}
// Driver Code
static void Main()
{
int[] seats = { 2, 3, 4, 5, 1 };
int k = seats.Length;
int n = 6;
Console.Write(maxProfit(seats, k, n));
}
}
// This code is contributed by divyeshrabadiya07 Javascript
C++
#include
using namespace std;
int maxProfit(int seats[],int k, int n)
{
sort(seats,seats+k);
int ans = 0;
int i = k - 1;
int j = k - 2;
while (n > 0 && j >= 0) {
if (seats[i] > seats[j]) {
ans = ans + min(n, (i - j)) * seats[i];
n = n - (i - j);
seats[i]--;
}
else {
while (j >= 0 && seats[j] == seats[i])
j--;
if (j < 0)
break;
ans = ans + min(n, (i - j)) * seats[i];
n = n - (i - j);
seats[i]--;
}
}
while (n > 0 && seats[i] != 0) {
ans = ans + min(n, k) * seats[i];
n -= k;
seats[i]--;
}
return ans;
}
int main()
{
int seats[] = { 2, 3, 4, 5, 1 };
int k = sizeof(seats) / sizeof(int);
int n = 6;
cout << maxProfit(seats, k, n);
return 0;
} 输出
22滑动窗口方法：
这个问题也可以使用滑动窗口技术来解决。
- 对于每个人，我们需要出售具有最高价格的票并将其价值减 1。
- 对阵列座位进行排序。
- 保持两个指针指向当前最大座位数和下一个最大座位数。
- 我们迭代直到我们的 n>0 并且数组中有第二大元素。
- 在每次迭代中，如果座位 [i] > 座位 [j] ，我们将座位 [i] 处的值，min(n, ij) 次添加到我们的答案中，并减少第 i 个索引处的值，否则我们找到 j 使得座位 [j] ]<座位[i]。如果没有这样的 j，我们就中断。
- 如果在迭代结束时我们的 n>0 和座位 [i]!=0，我们添加座位 [i] 直到 n>0 和座位 [i]!=0。
C++
#include
using namespace std;
int maxProfit(int seats[],int k, int n)
{
sort(seats,seats+k);
int ans = 0;
int i = k - 1;
int j = k - 2;
while (n > 0 && j >= 0) {
if (seats[i] > seats[j]) {
ans = ans + min(n, (i - j)) * seats[i];
n = n - (i - j);
seats[i]--;
}
else {
while (j >= 0 && seats[j] == seats[i])
j--;
if (j < 0)
break;
ans = ans + min(n, (i - j)) * seats[i];
n = n - (i - j);
seats[i]--;
}
}
while (n > 0 && seats[i] != 0) {
ans = ans + min(n, k) * seats[i];
n -= k;
seats[i]--;
}
return ans;
}
int main()
{
int seats[] = { 2, 3, 4, 5, 1 };
int k = sizeof(seats) / sizeof(int);
int n = 6;
cout << maxProfit(seats, k, n);
return 0;
}
输出
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