给定n个城市以及每对城市之间的距离，请选择k个城市来放置仓库(或ATM或Cloud Server)，以使城市到仓库(或ATM或Cloud Server)的最大距离最小。
例如，考虑以下四个城市，分别为0、1、2和3，以及它们之间的距离，如何在这四个城市中放置2个ATM，以使城市到ATM的最大距离最小。
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没有可用于该问题的多项式时间解，因为该问题是已知的NP-Hard问题。有一个多项式时间贪婪近似算法，贪婪算法提供的解决方案永远不会比最佳解决方案大两倍。仅当城市之间的距离遵循三角不等式(两个点之间的距离始终小于到第三个点的距离之和)时，贪婪的解决方案才有效。
2近似贪婪算法：
1)任意选择第一个中心。
2)使用以下条件选择剩余的k-1个中心。
令c1，c2，c3，…ci为已选择的中心。选择
(i + 1)’th中心，通过选择距离已经最远的城市
选定的中心，即具有最大跟随值的点p
Min [dist(p，c1)，dist(p，c2)，dist(p，c3)，…。 dist(p，ci)]
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示例(上图中的k = 3)
a)让第一个任意选择的顶点为0。
b)下一个顶点是1，因为1是距离0的最远顶点。
c)剩余的城市是2和3。计算它们与已选择的中心(0和1)的距离。贪婪算法基本上计算以下值。
从2到已经考虑的中心的所有距离中的最小值
Min [dist(2，0)，dist(2，1)] = Min [7，8] = 7
从3到已考虑的中心的所有距离中的最小值
Min [dist(3，0)，dist(3，1)] = Min [6，5] = 5
在计算完上述值之后，选择城市2，因为对应于2的值最大。
请注意，贪婪算法并未给出k = 2的最佳解，因为这只是一个近似算法，其边界是最佳值的两倍。
证明以上贪心算法为2近似值。
在最佳解决方案中，令OPT为城市到市中心的最大距离。我们需要证明，从贪婪算法获得的最大距离是2 * OPT。
证明可以使用矛盾来完成。
a)假设从最远点到所有中心的距离> 2·OPT。
b)这意味着所有中心之间的距离也> 2·OPT。
c)我们有k + 1个点，每对之间的距离> 2·OPT。
d)每个点都具有最佳解决方案的中心，并且距它的距离<= OPT。
e)在最优解中存在一对具有相同中心X的点(鸽子洞原理：k个最优中心，k + 1个点)
f)它们之间的距离最大为2·OPT(三角形不等式)，这是一个矛盾。
C++
// C++ program for the above approach
#include
using namespace std;
int maxindex(int* dist, int n)
{
int mi = 0;
for (int i = 0; i < n; i++) {
if (dist[i] > dist[mi])
mi = i;
}
return mi;
}
void selectKcities(int n, int weights[4][4], int k)
{
int* dist = new int[n];
vector centers;
for (int i = 0; i < n; i++) {
dist[i] = INT_MAX;
}
// index of city having the
// maximum distance to it's
// closest center
int max = 0;
for (int i = 0; i < k; i++) {
centers.push_back(max);
for (int j = 0; j < n; j++) {
// updating the distance
// of the cities to their
// closest centers
dist[j] = min(dist[j], weights[max][j]);
}
// updating the index of the
// city with the maximum
// distance to it's closest center
max = maxindex(dist, n);
}
// Printing the maximum distance
// of a city to a center
// that is our answer
cout << endl << dist[max] << endl;
// Printing the cities that
// were chosen to be made
// centers
for (int i = 0; i < centers.size(); i++) {
cout << centers[i] << " ";
}
cout << endl;
}
// Driver Code
int main()
{
int n = 4;
int weights[4][4] = { { 0, 4, 8, 5 },
{ 4, 0, 10, 7 },
{ 8, 10, 0, 9 },
{ 5, 7, 9, 0 } };
int k = 2;
// Function Call
selectKcities(n, weights, k);
}
// Contributed by Balu Nagar Java
// Java program for the above approach
import java.util.*;
class GFG{
static int maxindex(int[] dist, int n)
{
int mi = 0;
for(int i = 0; i < n; i++)
{
if (dist[i] > dist[mi])
mi = i;
}
return mi;
}
static void selectKcities(int n, int weights[][],
int k)
{
int[] dist = new int[n];
ArrayList centers = new ArrayList<>();
for(int i = 0; i < n; i++)
{
dist[i] = Integer.MAX_VALUE;
}
// Index of city having the
// maximum distance to it's
// closest center
int max = 0;
for(int i = 0; i < k; i++)
{
centers.add(max);
for(int j = 0; j < n; j++)
{
// Updating the distance
// of the cities to their
// closest centers
dist[j] = Math.min(dist[j],
weights[max][j]);
}
// Updating the index of the
// city with the maximum
// distance to it's closest center
max = maxindex(dist, n);
}
// Printing the maximum distance
// of a city to a center
// that is our answer
System.out.println(dist[max]);
// Printing the cities that
// were chosen to be made
// centers
for(int i = 0; i < centers.size(); i++)
{
System.out.print(centers.get(i) + " ");
}
System.out.print("\n");
}
// Driver Code
public static void main(String[] args)
{
int n = 4;
int[][] weights = new int[][]{ { 0, 4, 8, 5 },
{ 4, 0, 10, 7 },
{ 8, 10, 0, 9 },
{ 5, 7, 9, 0 } };
int k = 2;
// Function Call
selectKcities(n, weights, k);
}
}
// This code is contributed by nspatilme Python3
# Python3 program for the above approach
def maxindex(dist, n):
mi = 0
for i in range(n):
if (dist[i] > dist[mi]):
mi = i
return mi
def selectKcities(n, weights, k):
dist = [0]*n
centers = []
for i in range(n):
dist[i] = 10**9
# index of city having the
# maximum distance to it's
# closest center
max = 0
for i in range(k):
centers.append(max)
for j in range(n):
# updating the distance
# of the cities to their
# closest centers
dist[j] = min(dist[j], weights[max][j])
# updating the index of the
# city with the maximum
# distance to it's closest center
max = maxindex(dist, n)
# Printing the maximum distance
# of a city to a center
# that is our answer
# print()
print(dist[max])
# Printing the cities that
# were chosen to be made
# centers
for i in centers:
print(i, end = " ")
# Driver Code
if __name__ == '__main__':
n = 4
weights = [ [ 0, 4, 8, 5 ],
[ 4, 0, 10, 7 ],
[ 8, 10, 0, 9 ],
[ 5, 7, 9, 0 ] ]
k = 2
# Function Call
selectKcities(n, weights, k)
# This code is contributed by mohit kumar 29.5
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