POJ 2254 Globetrotter (计算几何 - 球面最短距离)
题目链接：POJ 2254
Description
As a member of an ACM programming team you'll soon find yourself always traveling around the world: Zürich, Philadelphia, San José, Atlanta,... from 1999 on the Contest Finals even will be on a different continent each year, so one day you might get to Japan or Australia.
At the contest site it would be interesting to know how many miles you are away from home. For this sake, your job is to write a program to compute the geographical distance between two given locations on the Earth's surface.
We assume that the Earth is a perfect sphere with a radius of exactly 6378 km. The geographical distance between A and B is the length of the geodetic line segment connecting A and B.
The geodetic line segment between two points on a sphere is the shortest connecting curve lying entirely in the surface of the sphere.
The value of pi is approximately 3.141592653589793.
Input
The input will consist of two parts: a list of cities and a list of queries.
City List
The city list consists of up to 100 lines, one line per city. Each line will contain a string ci and two real numbers lati and longi, representing the city name, its latitude and its longitude, respectively.
The city name will be shorter than 30 characters and will not contain white-space characters.
The latitude will be between -90 (South Pole) and +90 (North Pole). The longitude will be between -180 and +180 where negative numbers denote locations west of the meridian and positive numbers denote locations east of the meridian. (The meridian passes through Greenwich, London.)
The city list will be terminated by a line consisting of a single "#".
Query List
Each line will contain two city names A and B.
The query list will be terminated by the line "# #".
Output
For each query, print a line saying "A - B" where A and B are replaced by the city names. Then print a line saying x km" where x is replaced by the geographical distance (in km) between the two cities, rounded to the nearest integer.
If one of the cities in the query didn't occur in the city list, print a line saying "Unknown" instead. Print a blank line after each query.
Sample Input
Ulm 48.700 10.500
Freiburg 47.700 9.500
Philadelphia 39.883 -75.250
SanJose 37.366 -121.933
NorthPole 90 0
SouthPole -90 0
#
Ulm Philadelphia
Ulm SanJose
Freiburg Philadelphia
Freiburg SanJose
Ulm Freiburg
SanJose Philadelphia
Ulm LasVegas
Ulm Ulm
Ulm NorthPole
Ulm SouthPole
NorthPole SouthPole
# #
Sample Output
Ulm - Philadelphia
6536 km
Ulm - SanJose
9367 km
Freiburg - Philadelphia
6519 km
Freiburg - SanJose
9412 km
Ulm - Freiburg
134 km
SanJose - Philadelphia
4023 km
Ulm - LasVegas
Unknown
Ulm - Ulm
0 km
Ulm - NorthPole
4597 km
Ulm - SouthPole
15440 km
NorthPole - SouthPole
20037 km
Source
Solution
题意
给定一些城市的经纬度，然后给出若干个询问，每个询问包含两个城市，求这两个城市的球面最短距离。
思路
已知两点经纬度求球面最短距离的公式：
\(AB = R\cdot arccos(cos(wA)cos(wB)cos(jB-jA)+sin(wA)sin(wB))\)
其中 \(wA\) 和 \(jA\) 代表 \(A\) 的纬度和经度，\(wB\) 和 \(jB\) 代表 \(B\) 的纬度和经度。
Code
#include <iostream>
#include <cstdio>
#include <map>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
typedef double db;
const db pi = 3.141592653589793;
const db r = 6378.0;
class Point {
public:
double j, w;
Point(double j = 0, double w = 0) : j(j), w(w) {}
void input() {
scanf("%lf%lf", &w, &j);
w = w * pi / 180.0;
j = j * pi / 180.0;
}
db dis(Point a) {
return r * acos(cos(w) * cos(a.w) * cos(a.j - j) + sin(w) * sin(a.w));
}
};
map<string, Point> mp;
int main() {
string s1, s2;
while((cin >> s1) && s1[0] != '#') {
Point tmp;
tmp.input();
mp[s1] = tmp;
}
Point p1, p2;
while(cin >> s1 >> s2) {
if(s1[0] == '#' && s2[0] == '#') {
break;
}
cout << s1 << " - " << s2 << endl;
if(mp.find(s1) != mp.end() && mp.find(s2) != mp.end()) {
p1 = mp[s1], p2 = mp[s2];
double ans = p1.dis(p2);
printf("%.0lf km\n\n", ans);
} else {
printf("Unknown\n\n");
}
}
return 0;
}
