给定一个网格，该网格由大小为(N + 2)x(M + 2)的水平和垂直条以及两个数组H []和V []表示需要删除的水平和垂直条数组成，因此任务是找到删除一系列垂直和水平条时的最大面积。
例子：
Input: N = 3, M = 3, H[] = {2}, V[] = {2}
Output: 4
Explanation:
There are 3 bars in horizontal as well as vertical direction.
After removing bars the matrix looks like this:
Therefore, the largest area is 4.
Input: N = 3, M = 2, H[] = {1, 2, 3}, V[] = {1, 2}
Output: 12
方法：请按照以下步骤解决问题：
- 初始化两个集s1和s2以存储整数。
- 迭代范围[1，N + 1]并将每个整数存储在s1中。
- 类似地，在[1，M +1]范围内进行迭代，并将每个整数存储在s2中。
- 遍历数组H []并从s1中删除所有H [i] 。
- 同样，遍历数组V []并从s2中删除所有V [i] 。
- 将更新的s1和s2集转换为列表l1和l2。
- 以升序对两个列表进行排序。
- 遍历列表l1和l2 ，并将两个邻居之间的最大距离分别存储为maxH和maxV 。
- 将maxH和maxV的乘积打印为最大区域。
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 largest area
// when a series of horizontal &
// vertical bars are removed
void largestArea(int N, int M, int H[],
int V[], int h, int v)
{
// Stores all bars
set s1;
set s2;
// Insert horizontal bars
for (int i = 1; i <= N + 1; i++)
s1.insert(i);
// Insert vertictal bars
for (int i = 1; i <= M + 1; i++)
s2.insert(i);
// Remove horizontal separators from s1
for (int i = 0; i < h; i++) {
s1.erase(H[i]);
}
// Remove vertical separators from s2
for (int i = 0; i < v; i++) {
s2.erase(V[i]);
}
// Stores left out horizontal and
// vertical separators
int list1[s1.size()];
int list2[s2.size()];
int i = 0;
for (auto it1 = s1.begin(); it1 != s1.end(); it1++)
{
list1[i++] = *it1;
}
i = 0;
for (auto it2 = s2.begin(); it2 != s2.end(); it2++)
{
list2[i++] = *it2;
}
// Sort both list in
// ascending order
sort(list1, list1 + s1.size());
sort(list2, list2 + s2.size());
int maxH = 0, p1 = 0, maxV = 0, p2 = 0;
// Find maximum difference of neighbors of list1
for (int j = 0; j < s1.size(); j++) {
maxH = max(maxH, list1[j] - p1);
p1 = list1[j];
}
// Find max difference of neighbors of list2
for (int j = 0; j < s2.size(); j++) {
maxV = max(maxV, list2[j] - p2);
p2 = list2[j];
}
// Print largest volume
cout << (maxV * maxH) << endl;
}
// Driver code
int main()
{
// Given value of N & M
int N = 3, M = 3;
// Given arrays
int H[] = { 2 };
int V[] = { 2 };
int h = sizeof(H) / sizeof(H[0]);
int v = sizeof(V) / sizeof(V[0]);
// Function call to find the largest
// area when a series of horizontal &
// vertical bars are removed
largestArea(N, M, H, V, h, v);
return 0;
}
// This code is contributed by divyeshrabadiya07. Java
// Java program for the above approach
import java.lang.*;
import java.util.*;
class GFG {
// Function to find the largest area
// when a series of horizontal &
// vertical bars are removed
static void largestArea(int N, int M,
int[] H, int[] V)
{
// Stores all bars
Set s1 = new HashSet<>();
Set s2 = new HashSet<>();
// Insert horizontal bars
for (int i = 1; i <= N + 1; i++)
s1.add(i);
// Insert vertictal bars
for (int i = 1; i <= M + 1; i++)
s2.add(i);
// Remove horizontal separators from s1
for (int i = 0; i < H.length; i++) {
s1.remove(H[i]);
}
// Remove vertical separators from s2
for (int i = 0; i < V.length; i++) {
s2.remove(V[i]);
}
// Stores left out horizontal and
// vertical separators
int[] list1 = new int[s1.size()];
int[] list2 = new int[s2.size()];
int i = 0;
Iterator it1 = s1.iterator();
while (it1.hasNext()) {
list1[i++] = (int)it1.next();
}
i = 0;
Iterator it2 = s2.iterator();
while (it2.hasNext()) {
list2[i++] = (int)it2.next();
}
// Sort both list in
// ascending order
Arrays.sort(list1);
Arrays.sort(list2);
int maxH = 0, p1 = 0, maxV = 0, p2 = 0;
// Find maximum difference of neighbors of list1
for (int j = 0; j < list1.length; j++) {
maxH = Math.max(maxH, list1[j] - p1);
p1 = list1[j];
}
// Find max difference of neighbors of list2
for (int j = 0; j < list2.length; j++) {
maxV = Math.max(maxV, list2[j] - p2);
p2 = list2[j];
}
// Print largest volume
System.out.println(maxV * maxH);
}
// Driver Code
public static void main(String[] args)
{
// Given value of N & M
int N = 3, M = 3;
// Given arrays
int[] H = { 2 };
int[] V = { 2 };
// Function call to find the largest
// area when a series of horizontal &
// vertical bars are removed
largestArea(N, M, H, V);
}
} Python3
# Python 3 program for the above approach
# Function to find the largest area
# when a series of horizontal &
# vertical bars are removed
def largestArea(N, M, H,
V, h, v):
# Stores all bars
s1 = set([]);
s2 = set([]);
# Insert horizontal bars
for i in range(1, N + 2):
s1.add(i);
# Insert vertictal bars
for i in range(1, M + 2):
s2.add(i);
# Remove horizontal separators from s1
for i in range(h):
s1.remove(H[i]);
# Remove vertical separators from s2
for i in range( v ):
s2.remove(V[i]);
# Stores left out horizontal and
# vertical separators
list1 = [0] * len(s1)
list2 = [0]*len(s2);
i = 0;
for it1 in s1:
list1[i] = it1;
i += 1
i = 0;
for it2 in s2:
list2[i] = it2
i += 1
# Sort both list in
# ascending order
list1.sort();
list2.sort();
maxH = 0
p1 = 0
maxV = 0
p2 = 0;
# Find maximum difference of neighbors of list1
for j in range(len(s1)):
maxH = max(maxH, list1[j] - p1);
p1 = list1[j];
# Find max difference of neighbors of list2
for j in range(len(s2)):
maxV = max(maxV, list2[j] - p2);
p2 = list2[j];
# Print largest volume
print((maxV * maxH))
# Driver code
if __name__ == "__main__":
# Given value of N & M
N = 3
M = 3;
# Given arrays
H = [2]
V = [2];
h = len(H)
v = len(V);
# Function call to find the largest
# area when a series of horizontal &
# vertical bars are removed
largestArea(N, M, H, V, h, v);
# This code is contributed by ukasp.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to find the largest area
// when a series of horizontal &
// vertical bars are removed
static void largestArea(int N, int M,
int[] H, int[] V)
{
// Stores all bars
HashSet s1 = new HashSet();
HashSet s2 = new HashSet();
// Insert horizontal bars
for (int i = 1; i <= N + 1; i++)
s1.Add(i);
// Insert vertictal bars
for (int i = 1; i <= M + 1; i++)
s2.Add(i);
// Remove horizontal separators from s1
for (int i = 0; i < H.Length; i++) {
s1.Remove(H[i]);
}
// Remove vertical separators from s2
for (int i = 0; i < V.Length; i++) {
s2.Remove(V[i]);
}
// Stores left out horizontal and
// vertical separators
int[] list1 = new int[s1.Count];
int[] list2 = new int[s2.Count];
int I = 0;
foreach(int it1 in s1)
{
list1[I++] = it1;
}
I = 0;
foreach(int it2 in s2)
{
list2[I++] = it2;
}
// Sort both list in
// ascending order
Array.Sort(list1);
Array.Sort(list2);
int maxH = 0, p1 = 0, maxV = 0, p2 = 0;
// Find maximum difference of neighbors of list1
for (int j = 0; j < list1.Length; j++) {
maxH = Math.Max(maxH, list1[j] - p1);
p1 = list1[j];
}
// Find max difference of neighbors of list2
for (int j = 0; j < list2.Length; j++) {
maxV = Math.Max(maxV, list2[j] - p2);
p2 = list2[j];
}
// Print largest volume
Console.WriteLine(maxV * maxH);
}
// Driver code
static void Main()
{
// Given value of N & M
int N = 3, M = 3;
// Given arrays
int[] H = { 2 };
int[] V = { 2 };
// Function call to find the largest
// area when a series of horizontal &
// vertical bars are removed
largestArea(N, M, H, V);
}
}
// This code is contributed by divyesh072019. 输出：
4时间复杂度： max(N，M)*(log(max(N，M)))
辅助空间：O(max(N，M))