您将获得一个2D网格，其值为0或1，其中每个1表示组中某人的家。两个或两个以上的人希望见面并最大程度地缩短总行驶距离。他们可以在任何地方见面，这意味着可能没有家。
- 使用曼哈顿距离计算距离，其中距离(p1，p2)= | p2.x – p1.x | + | p2.y – p1.y |。
- 找到达到最佳集合点所需的总距离(总距离最小)。
例子：
Input : grid[][] = {{1, 0, 0, 0, 1},
{0, 0, 0, 0, 0},
{0, 0, 1, 0, 0}};
Output : 6
Best meeting point is (0, 2).
Total distance traveled is 2 + 2 + 2 = 6
Input : grid[3][5] = {{1, 0, 1, 0, 1},
{0, 1, 0, 0, 0},
{0, 1, 1, 0, 0}};
Output : 11
脚步 ：-
1)存储所有组成员的所有水平和垂直位置。
2)现在将其排序以找到最小的中间位置，这将是最佳的交汇点。
3)找到所有成员与最佳集合点的距离。
例如，在上图中，水平位置为{0，2，0}，垂直位置为{0，2，4}。对两者进行排序后，我们得到{0，0，2}和{0，2，4}。中间点是(0，2)。
注意：即使没有。的1有两个中间点，那么它也起作用。两个中间点意味着它总是有两个最佳会合点。两种情况将给出相同的距离。因此，我们将仅考虑一个最佳的交汇点以避免更多的开销，因为我们的目标是仅查找距离。
C++
/* C++ program to find best meeting point in 2D array*/
#include
using namespace std;
#define ROW 3
#define COL 5
int minTotalDistance(int grid[][COL]) {
if (ROW == 0 || COL == 0)
return 0;
vector vertical;
vector horizontal;
// Find all members home's position
for (int i = 0; i < ROW; i++) {
for (int j = 0; j < COL; j++) {
if (grid[i][j] == 1) {
vertical.push_back(i);
horizontal.push_back(j);
}
}
}
// Sort positions so we can find most
// beneficial point
sort(vertical.begin(),vertical.end());
sort(horizontal.begin(),horizontal.end());
// middle position will always beneficial
// for all group members but it will be
// sorted which we have alredy done
int size = vertical.size()/2;
int x = vertical[size];
int y = horizontal[size];
// Now find total distance from best meeting
// point (x,y) using Manhattan Distance formula
int distance = 0;
for (int i = 0; i < ROW; i++)
for (int j = 0; j < COL; j++)
if (grid[i][j] == 1)
distance += abs(x - i) + abs(y - j);
return distance;
}
// Driver program to test above functions
int main() {
int grid[ROW][COL] = {{1, 0, 1, 0, 1}, {0, 1, 0, 0, 0},{0, 1, 1, 0, 0}};
cout << minTotalDistance(grid);
return 0;
} Java
/* Java program to find best
meeting point in 2D array*/
import java.util.*;
class GFG
{
static int ROW = 3;
static int COL =5 ;
static int minTotalDistance(int grid[][])
{
if (ROW == 0 || COL == 0)
return 0;
Vector vertical = new Vector();
Vector horizontal = new Vector();
// Find all members home's position
for (int i = 0; i < ROW; i++)
{
for (int j = 0; j < COL; j++)
{
if (grid[i][j] == 1)
{
vertical.add(i);
horizontal.add(j);
}
}
}
// Sort positions so we can find most
// beneficial point
Collections.sort(vertical);
Collections.sort(horizontal);
// middle position will always beneficial
// for all group members but it will be
// sorted which we have alredy done
int size = vertical.size() / 2;
int x = vertical.get(size);
int y = horizontal.get(size);
// Now find total distance from best meeting
// point (x,y) using Manhattan Distance formula
int distance = 0;
for (int i = 0; i < ROW; i++)
for (int j = 0; j < COL; j++)
if (grid[i][j] == 1)
distance += Math.abs(x - i) +
Math.abs(y - j);
return distance;
}
// Driver code
public static void main(String[] args)
{
int grid[][] = {{1, 0, 1, 0, 1},
{0, 1, 0, 0, 0},
{0, 1, 1, 0, 0}};
System.out.println(minTotalDistance(grid));
}
}
// This code is contributed by 29AjayKumar Python3
# Python program to find best meeting point in 2D array
ROW = 3
COL = 5
def minTotalDistance(grid: list) -> int:
if ROW == 0 or COL == 0:
return 0
vertical = []
horizontal = []
# Find all members home's position
for i in range(ROW):
for j in range(COL):
if grid[i][j] == 1:
vertical.append(i)
horizontal.append(j)
# Sort positions so we can find most
# beneficial point
vertical.sort()
horizontal.sort()
# middle position will always beneficial
# for all group members but it will be
# sorted which we have alredy done
size = len(vertical) // 2
x = vertical[size]
y = horizontal[size]
# Now find total distance from best meeting
# point (x,y) using Manhattan Distance formula
distance = 0
for i in range(ROW):
for j in range(COL):
if grid[i][j] == 1:
distance += abs(x - i) + abs(y - j)
return distance
# Driver Code
if __name__ == "__main__":
grid = [[1, 0, 1, 0, 1],
[0, 1, 0, 0, 0],
[0, 1, 1, 0, 0]]
print(minTotalDistance(grid))
# This code is contributed by
# sanjeev2552C#
/* C# program to find best
meeting point in 2D array*/
using System;
using System.Collections.Generic;
class GFG
{
static int ROW = 3;
static int COL = 5 ;
static int minTotalDistance(int [,]grid)
{
if (ROW == 0 || COL == 0)
return 0;
List vertical = new List();
List horizontal = new List();
// Find all members home's position
for (int i = 0; i < ROW; i++)
{
for (int j = 0; j < COL; j++)
{
if (grid[i, j] == 1)
{
vertical.Add(i);
horizontal.Add(j);
}
}
}
// Sort positions so we can find most
// beneficial point
vertical.Sort();
horizontal.Sort();
// middle position will always beneficial
// for all group members but it will be
// sorted which we have alredy done
int size = vertical.Count / 2;
int x = vertical[size];
int y = horizontal[size];
// Now find total distance from best meeting
// point (x,y) using Manhattan Distance formula
int distance = 0;
for (int i = 0; i < ROW; i++)
for (int j = 0; j < COL; j++)
if (grid[i, j] == 1)
distance += Math.Abs(x - i) +
Math.Abs(y - j);
return distance;
}
// Driver code
public static void Main(String[] args)
{
int [,]grid = {{1, 0, 1, 0, 1},
{0, 1, 0, 0, 0},
{0, 1, 1, 0, 0}};
Console.WriteLine(minTotalDistance(grid));
}
}
// This code is contributed by PrinciRaj1992 输出：
11
时间复杂度： O(M * N)
辅助空间： O(N)