给定整数N,L和W表示行数以及每行中房屋的长度和宽度,以及代表每个房屋高度的数组Heights [] ,任务是找到所需的外墙总面积为如果相邻的房屋共用一堵墙,则进行绘制。
例子:
Input: N = 4, W = 1, L = 1, Heights[] = {1, 2, 3, 4}
Output: 28
Explanation:
The area of house-1 painted = 2 * 1 + 1 * 1 = 3
The area of house-2 painted = 2 * 2 + 1 * 1 = 5
The area of house-3 painted = 2 * 3 + 1 * 1 = 7
The area of house-4 painted = 2 * 4 + 1 * 1 + 1 * 4 = 13
Therefore, the total area painted = 28
Input: N = 7, W = 1, L = 1, Heights[] = {4, 3, 1, 2, 3, 4, 2}
Output: 52
方法:可以使用贪婪技术解决问题。请按照以下步骤解决问题:
- 涂漆的第一栋房屋的墙壁总面积等于2 * w * h 1 + l *(h 1 + max(0,h 2 – h 1 ))。
- 最后涂漆的房屋墙壁的总面积等于2 * w * h n + l *(h n + max(0,h n – h n-1 ))。
- 除了第一个和最后一个房屋外,粉刷过的房屋墙壁的总面积等于2 * w * h i + l *(max(0,h i – h i + 1 )+ max(0,h i – h i-1 ))。
- 因此,将使用以下公式计算所有粉刷房屋的总面积:
Total area printed = 2 * w(h1 + …+hn) + l(h1 + hn) + l * Σabs(hi – hi-1).
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the total area
// of walls painted in N row-houses
void areaToPaint(int N, int W, int L,
int Heights[])
{
// Stores total area of N row-houses
// that needs to be painted
int total = 0;
// Traverse the array of wall heights
for (int i = 0; i < N; i++) {
// Update total area painted
total += 2 * Heights[i] * W;
}
// Update total
total += L * (Heights[0]
+ Heights[N - 1]);
// Traverse all the houses and
// print the shared walls
for (int i = 1; i < N; i++) {
// Update total
total += L * abs(Heights[i]
- Heights[i - 1]);
}
// Print total area needs to paint
cout << total;
}
// Driver Code
int main()
{
// Given N, W & L
int N = 7, W = 1, L = 1;
// Given heights of houses
int Heights[N]
= { 4, 3, 1, 2, 3, 4, 2 };
// Function Call
areaToPaint(N, W, L, Heights);
}
Java
// Java code of above approach
import java.util.*;
class GFG {
// Function to find the total area
// of walls painted in N row-houses
static void areaToPaint(int N, int W, int L,
int Heights[])
{
// Stores total area of N row-houses
// that needs to be painted
int total = 0;
// Traverse the array of wall heights
for (int i = 0; i < N; i++) {
// Update total area painted
total += 2 * Heights[i] * W;
}
// Update total
total += L * (Heights[0]
+ Heights[N - 1]);
// Traverse all the houses and
// print the shared walls
for (int i = 1; i < N; i++) {
// Update total
total += L * Math.abs(Heights[i]
- Heights[i - 1]);
}
// Print total area needs to paint
System.out.print(total);
}
// Driver code
public static void main(String[] args)
{
// Given N, W & L
int N = 7, W = 1, L = 1;
// Given heights of houses
int Heights[]
= { 4, 3, 1, 2, 3, 4, 2 };
// Function Call
areaToPaint(N, W, L, Heights);
}
}
// This code is contributed by offbeat
Python3
# Python 3 program for the above approach
# Function to find the total area
# of walls painted in N row-houses
def areaToPaint(N, W, L, Heights):
# Stores total area of N row-houses
# that needs to be painted
total = 0
# Traverse the array of wall heights
for i in range(N):
# Update total area painted
total += 2 * Heights[i] * W
# Update total
total += L * (Heights[0]
+ Heights[N - 1])
# Traverse all the houses and
# print the shared walls
for i in range(1, N):
# Update total
total += L * abs(Heights[i]
- Heights[i - 1])
# Print total area needs to paint
print(total)
# Driver Code
if __name__ == "__main__":
# Given N, W & L
N = 7
W = 1
L = 1
# Given heights of houses
Heights = [4, 3, 1, 2, 3, 4, 2]
# Function Call
areaToPaint(N, W, L, Heights)
# This code is contributed by chitranayal.
输出:
52
时间复杂度: O(N)
辅助空间: O(N)