BOJ1018 -- Airline Hub_world wide flyer has landing rights at several air-优快云博客

本文链接：https://blog.youkuaiyun.com/wenlele/article/details/6527294

本文介绍了一个关于寻找全球范围内最佳航空枢纽位置的问题。通过计算不同机场间的球面距离，找到那个能够最小化最大直接飞行距离的机场作为中枢。文章提供了一段C++代码实现这一计算过程。

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Airline HubTime Limit: 1000MS Memory Limit: 65536K

Description
World Wide Flyer has landing rights at several airports throughout the world. They wish to place their central hub at the airport that minimizes the maximum direct flying distance from the hub to any other airport in the world.

Input
Input consists of a line containing n <= 1000, the number of airports. n lines follow, each giving the latitude (between -90 and +90 degrees) and longitude (between -180 and +180 degrees) of an airport.

Output
To two decimal places, give the latitude and longitude of the airport that best serves as a hub. If there are several any one will do.

Sample Input

3
3.2 -15.0
20.1 -175
-30.2 10

Sample Output

3.20 -15.00

Source
Waterloo local 2000.01.29

解：

一首水题，主要是要懂球面的距离计算。

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;

typedef struct airport{
   double x;
   double y;   
}A;

double distance(double x1, double y1, double x2, double y2)
{
   double n1 = x1 - x2;
   double n2 = y1 - y2;
   return sqrt(n1*n1 + n2*n2);
}

A a[1010];
double d[1010][1010];

int main()
{      
   int n, minNum, tempNum;
   double min = 0.0, temp = 0.0;
   int count = 0;
   
   cin >> n;
   for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
         d[i][j] = 0.0; 
   
   for (int i = 0; i < n; i++){
      cin >> a[i].x >> a[i].y;
   }
   
   for (int i = 0; i < n; i++)
      for (int j = i+1; j < n; j++){
         d[j][i] = d[i][j] = distance(a[i].x, a[i].y, a[j].x, a[j].y);
      }
   
   for (int i = 0; i < n; i++){      
      for (int j = 0; j < n; j++)
         printf("%f/t", d[i][j]);
      cout << endl;
   }
   
   for (int i = 0; i < n; i++)
      min += d[0][i];
   minNum = 0;
   cout << min << endl;
   for (int i = 1; i < n; i++){
      temp = 0.0;
      for (int j = 0; j < n; j++)
         temp += d[i][j];
      cout << temp << endl;
      if (min > temp){
         minNum = i;
         min = temp;
      }
   }   
   printf("%.2f %.2f/n", a[minNum].x, a[minNum].y);
   
   system("pause");
   return 0;
}