UVa 10075 - Airlines

题目：给出地球上一些点的经度和纬度，以及地点之间是否直达的航班，然后询问两地间最短的行程。

分析：计算几何、最短路、字符串、hash。

首先，利用hash建立地名（单词）和编号的映射；

然后，将大地坐标转换，求出有直达航班地点之间的球面最短距离；

d = r*sqrt(2-2*(cos(lat1)*cos(lat2)*cos(lon1-lon2)+sin(lat1)*sin(lat2)) （推导见11817）

最后，利用floyd计算所有点间的最短距离，查询输出即可。

注意：1.有向图；2.距离取整要在计算最短路之前（四舍五入）；

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;

//hash__begin
typedef struct hnode
{
	int    id;
	char   name[ 21 ];
	hnode* next;
}hnode;
hnode* List[ 257 ];
hnode  Node[ 105 ];
int table[20]={	13,17,31,71,131,171,313,717,1313,1717,
				3131,7171,13131,17171,31313,71717,131313,
				171717,313131,717171};

class hash {
	int size;
public:
	hash() {
		size = 0;
		memset( List, 0, sizeof(List) );
	}
	//计算hash值 
	int calc( char* str ) {
		int value = 0;
		for ( int i = 0 ; str[i] ; ++ i )
			value = (value+str[i]*table[i])%257;
		return value;
	}
	//插入 
	void insert( char* str, int id ) {
		int value = calc(str);
		Node[size].next = List[value];  
    	Node[size].id   = id;
    	List[value] = &Node[size];  
    	strcpy(Node[size ++].name,str);
	}
	//查询 
	int find( char* str ) {
		int value = calc(str);
		for ( hnode* p = List[value] ; p ; p = p->next )
			if ( !strcmp( str, p->name ) )
				return p->id;
	}
};
//hash__end

char   name[105][21];
char   city1[21],city2[21];
double lat[105],lon[105],dis;
int    path[105][105];

//大地坐标转化 
int dist( double l1, double d1, double l2, double d2 )
{
	double r = 6378;
	double p = 3.141592653589793;
	l1 *= p/180.0; 
	l2 *= p/180.0;
	d1 *= p/180.0; 
	d2 *= p/180.0;
	double d = r*sqrt(2-2*(cos(l1)*cos(l2)*cos(d1-d2)+sin(l1)*sin(l2)));
	return (int)(0.5+2*asin(d/(2*r))*r);	
}

//floyd计算多元最短路 
void floyd( int n )
{
	for ( int k = 0 ; k < n ; ++ k )
	for ( int i = 0 ; i < n ; ++ i )
	for ( int j = 0 ; j < n ; ++ j )
	    if ( path[i][k] != -1 && path[k][j] != -1 )
	    if ( path[i][j] == -1 || path[i][j] > path[i][k] + path[k][j] )
	    	path[i][j] = path[i][k] + path[k][j];
}

int main()
{
	int N,M,Q,id1,id2,T = 1;
	while ( ~scanf("%d%d%d",&N,&M,&Q) && N+M+Q ) {
		if ( T > 1 ) printf("\n");
		printf("Case #%d\n",T ++);
		
		//输入数据到hash表 
		hash Hash;
		for ( int i = 0 ; i < N ; ++ i ) {
			scanf("%s%lf%lf",name[i],&lat[i],&lon[i]);
			Hash.insert( name[i], i );
		}
		
		//初始化距离，只计算可直达 
		for ( int i = 0 ; i < N ; ++ i )
		for ( int j = 0 ; j < N ; ++ j )
			path[i][j] = -1;
		for ( int i = 0 ; i < M ; ++ i ) {
			scanf("%s%s",city1,city2);
			id1 = Hash.find( city1 );
			id2 = Hash.find( city2 );
			path[id1][id2] = dist( lat[id1], lon[id1], lat[id2], lon[id2] );
		}
		
		//计算有向图多元最短路 
		floyd( N );
		
		//查询输出 
		for ( int i = 0 ; i < Q ; ++ i ) {
			scanf("%s%s",city1,city2);
			id1 = Hash.find( city1 );
			id2 = Hash.find( city2 );
			if ( path[id1][id2] == -1 )
				printf("no route exists\n");
			else printf("%d km\n",path[id1][id2]);
		}
	}
	return 0;
}