给定两个数组A[]和B[]由N 个整数和一个整数K 组成，任务是通过以下操作最大化从数组A[]计算的总和：
- 对于B[] 中包含0 的每个索引，将A[]中的相应索引添加到总和中。
- 对于B[] 中包含1 的每个索引，将A[]中相应索引处的值添加到最多K 个此类索引的总和中。对于剩余的索引，从总和中减去。
例子：
Input: A[] = {5, 4, 6, 2, 8}, B[] = {1, 0, 1, 1, 0}, K = 2
Output: 21
Explanation:
Add A[1] and A[4] to the sum as B[1] = B[4] = 0
Therefore, sum = 4 + 8 = 12.
Now, add A[0] and A[3] to the sum as K elements can be added.
Finally, subtract 2 from the sum.
Therefore, the maximum possible sum = 12 + 5 + 6 – 2 = 21
Input: A[] = {5, 2, 1, 8, 10, 5}, B[] = {1, 1, 1, 1, 0, 0}, K = 3
Output: 29
方法：
请按照以下步骤解决问题：
- 按降序对数组 A[] 进行排序。
- 要使总和最大化，请添加与B[]中的索引包含1对应的已排序数组中的前K 个元素。减去剩余的此类元素。
- 将A[] 中与B[] 中包含0的索引对应的所有值加到总和中。
下面是上述方法的实现：
C++
// C++ Program to maximise the
// sum of the given array
#include
using namespace std;
// Comparator to sort the array
// in ascending order
bool compare(pair p1,
pair p2)
{
return p1.first > p2.first;
}
// Function to maximise the sum of
// the given array
int maximiseScore(int A[], int B[],
int K, int N)
{
// Stores {A[i], B[i]} pairs
vector > pairs(N);
for (int i = 0; i < N; i++) {
pairs[i] = make_pair(A[i], B[i]);
}
// Sort in descending order of the
// values in the array A[]
sort(pairs.begin(), pairs.end(), compare);
// Stores the maximum sum
int sum = 0;
for (int i = 0; i < N; i++) {
// If B[i] is equal to 0
if (pairs[i].second == 0) {
// Simply add A[i] to the sum
sum += pairs[i].first;
}
else if (pairs[i].second == 1) {
// Add the highest K numbers
if (K > 0) {
sum += pairs[i].first;
K--;
}
// Subtract from the sum
else {
sum -= pairs[i].first;
}
}
}
// Return the sum
return sum;
}
// Driver Code
int main()
{
int A[] = { 5, 4, 6, 2, 8 };
int B[] = { 1, 0, 1, 1, 0 };
int K = 2;
int N = sizeof(A) / sizeof(int);
cout << maximiseScore(A, B, K, N);
return 0;
} Java
// Java program to maximise the
// sum of the given array
import java.util.*;
class Pair implements Comparable
{
int first, second;
Pair(int x, int y)
{
first = x;
second = y;
}
public int compareTo(Pair p)
{
return p.first - first;
}
}
class GFG{
// Function to maximise the sum of
// the given array
static int maximiseScore(int A[], int B[],
int K, int N)
{
// Stores {A[i], B[i]} pairs
ArrayList pairs = new ArrayList<>();
for(int i = 0; i < N; i++)
{
pairs.add(new Pair(A[i], B[i]));
}
// Sort in descending order of the
// values in the array A[]
Collections.sort(pairs);
// Stores the maximum sum
int sum = 0;
for(int i = 0; i < N; i++)
{
// If B[i] is equal to 0
if (pairs.get(i).second == 0)
{
// Simply add A[i] to the sum
sum += pairs.get(i).first;
}
else if (pairs.get(i).second == 1)
{
// Add the highest K numbers
if (K > 0)
{
sum += pairs.get(i).first;
K--;
}
// Subtract from the sum
else
{
sum -= pairs.get(i).first;
}
}
}
// Return the sum
return sum;
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 5, 4, 6, 2, 8 };
int B[] = { 1, 0, 1, 1, 0 };
int K = 2;
int N = A.length;
System.out.print(maximiseScore(A, B, K, N));
}
}
// This code is contributed by jrishabh99 Python3
# Python Program to maximise the
# sum of the given array
# Comparator to sort the array
# in ascending order
def compare(p1, p2):
return p1[0] > p2[0]
# Function to maximise the sum of
# the given array
def maximiseScore(A, B, K, N):
# Stores {A[i], B[i]} pairs
pairs = []
for i in range(N):
pairs.append([A[i], B[i]])
# Sort in descending order of the
# values in the array A[]
pairs.sort(key = lambda x:x[0], reverse = True)
# Stores the maximum sum
Sum = 0
for i in range(N):
# If B[i] is equal to 0
if(pairs[i][1] == 0):
# Simply add A[i] to the sum
Sum += pairs[i][0]
elif(pairs[i][1] == 1):
# Add the highest K numbers
if(K > 0):
Sum += pairs[i][0]
K -= 1
# Subtract from the sum
else:
Sum -= pairs[i][0]
# Return the sum
return Sum
# Driver Code
A = [5, 4, 6, 2, 8]
B = [1, 0, 1, 1, 0]
K = 2
N = len(A)
print(maximiseScore(A, B, K, N))
# This code is contributed by avanitrachhadiya2155输出：
21时间复杂度： O(N*log(N))
辅助空间： O(N)
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