学步园

    原题URL:http://poj.org/problem?id=2412；

    这道题是一道与空间几何相关的题目，完全数学题。强调几点：

    1.两个城市位于同一位置时，也就是平面的法向量为0时，不可用向量的夹角公式求解两向量的夹角。

     2.四舍五入的一个小技巧，原浮点数加0.5之后取整。

     这道题我写了两个版本，一个是用了STL的find函数，自己重载了==运算符，另一个是自己写了个查找函数。发现泛型还是要慢一些。下面是我的AC代码，和大家分享一下，欢迎拍砖，共同进步。

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

const int Max = 20 + 10;
const int CityMax = 110;
const double PI = 3.1415926;
const double Radius = 6378.0;

struct City
{
	char name[Max];
	double latitude, longitude;
	double x, y, z;

	friend bool operator == (const City & a, const City &b)
	{
		if(!strcmp(a.name, b.name)) return true;
		else return false;
	}
};

City c[CityMax];

struct Vector
{
	double x, y, z;
};

double radian(double d)
{
	return d * PI / 180.0;
}

double square(double d)
{
	return d * d;
}

double angle(Vector &v, City &c)
{
	return acos((v.x * c.x + v.y * c.y + v.z * c.z) / sqrt( square(v.x) + square(v.y) + square(v.z) ) / sqrt( square(c.x) + square(c.y) + square(c.z) ));
}

City * findCity(City *first, City *end, City &c)
{
	while(first != end)
	{
		if(!strcmp(first->name, c.name)) return first;
		first ++;
	}
	return first;
}

int main()
{
	int index = 0;
	City A, B, C;

	while(scanf("%s", c[index].name) && strcmp(c[index].name, "#") != 0)
	{
		scanf("%lf %lf", &c[index].latitude, &c[index].longitude);

		c[index].x = Radius * cos(radian(c[index].latitude)) * sin(radian(c[index].longitude));
		c[index].y = Radius * cos(radian(c[index].latitude)) * cos(radian(c[index].longitude));
		c[index].z = Radius * sin(radian(c[index].latitude));

		index ++;
	}
	
	Vector vec;
	double ang, dist;
	while(scanf("%s", A.name) && strcmp(A.name, "#"))
	{
		scanf("%s %s", B.name, C.name);

			City * a, * b, *cen;
			//a = findCity(c, c+index, A), b = findCity(c, c+index, B), cen = findCity(c, c+index, C);//调用函数0MS通过
			a = find(c, c+index, A), b = find(c, c+index, B), cen = find(c, c+index, C); //泛型花了16MS
			if(a == c+index || b == c+index || cen == c+index)
			{
				printf("%s is ? km off %s/%s equidistance.\n", C.name, A.name, B.name);
			}
			else
			{
				vec.x = a->x - b->x, vec.y = a->y - b->y, vec.z = a->z - b->z;
				if(vec.x == 0.0 && vec.y == 0.0 && vec.z == 0.0) dist = 0;
				else 
				{
					ang = fabs(PI/2 - angle(vec, *cen));
					dist = ang*Radius + 0.5;
				}
				printf("%s is %d km off %s/%s equidistance.\n", C.name, (int)(dist), A.name, B.name);
			}
	}

	system("pause");
	return 0;
}