地铁换乘—华为2014校招机试样题 —Dijkstra 和 Floyd-Warshall 算法解决

地铁换乘——华为2014校招机试样题

——方法一：Dijkstra最短路径算法

 

原题如下：

地铁换乘

描述:  

已知2条地铁线路，其中A为环线，B为东西向线路，线路都是双向的。经过的站点名分别如下，两条线交叉的换乘点用T1、T2表示。编写程序，任意输入两个站点名称，输出乘坐地铁最少需要经过的车站数量（含输入的起点和终点，换乘站点只计算一次）。

地铁线A（环线）经过车站：A1 A2 A3 A4 A5 A6 A7 A8 A9 T1 A10A11 A12 A13 T2 A14 A15 A16 A17 A18

地铁线B（直线）经过车站：B1 B2 B3 B4 B5 T1 B6 B7 B8 B9 B10T2 B11 B12 B13 B14 B15

运行时间限制:   无限制

内存限制:   无限制

输入:   输入两个不同的站名

输出:   输出最少经过的站数,含输入的起点和终点，换乘站点只计算一次

样例输入:   A1 A3

样例输出:   3

 

上星期试着解这道题，当时还没看数据结构图的部分，想着用双链表加上各种条件判断来解决，发现太复杂了，遂网上搜了搜，这里提供了一种通过图来解决的Floyed算法，链接如下：

http://blog.csdn.net/arcsinsin/article/details/11267755

鉴于此，这几天赶紧去看了下图的部分，刚看了Dijkstra单源点到所有终点的最短路径算法，想来可以改造下解决这里的问题，遂解决如下。

关于Dijkstra算法可以参见这里：

http://www.cnblogs.com/dolphin0520/archive/2011/08/26/2155202.html

 

#include 
#include 
#include 
using namespace std;

#define SIZE_A 21
#define SIZE_B 17

#define N 35

// 20 + 1
string A[] = {"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1","A10","A11","A12","A13","T2","A14","A15","A16","A17","A18","A1"};
// 17
string B[] = {"B1","B2","B3","B4","B5","T1","B6","B7","B8","B9","B10","T2","B11","B12","B13","B14","B15"}; 
// 35
string Node[] = {"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1","A10","A11","A12","A13","T2","A14","A15","A16","A17","A18",
				"B1","B2","B3","B4","B5","B6","B7","B8","B9","B10","B11","B12","B13","B14","B15"}; 

struct Graph {
	int matrix[N][N];
	int n;
	int e;
};

Graph g;
bool s[N];
int dist[N];
int path[N];

int length=0;

string StrBegin,StrEnd;

void Dijkstra(int v0, int v1);

int GetPos(string *array, string & str)
{
	int n=0;
	if (str[0] == 'B') {
		n+=20;
		array+=20;
	}
	if (str == "T1") return 9;
	if (str == "T2") return 14;

	while (*array != str) {
		array ++;
		n++;
	}
	return n;
}

void BuildGraph()
{
	// 初始化matrix
	for (int i=0; i s;
	//while (v1 != v0)
	//{
	//	s.push(v1);
	//	v1 = path[v1];
	//}
	//s.push(v0);
	//while (!s.empty())
	//{
	//	length++;
	//	if (1)	// debug
	//	{
	//		int pos = s.top();
	//		string str = Node[pos];
	//		str += "\0";
	//		cout<>StrBegin>>StrEnd;
		if (StrBegin == StrEnd)
			cout<<"1"<

 ——方法二：Floyd-Warshall顶点对最短路径算法

该算法可以求出各顶点对之间的最短路径，写法简单，主要核心就是一个三层嵌套循环即可。

#include 
#include 
#include 
using namespace std;

#define SIZE_A 21
#define SIZE_B 17

#define N 35
#define INF 0xfffff

// 20 + 1
string A[] = {"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1","A10","A11","A12","A13","T2","A14","A15","A16","A17","A18","A1"};
// 17
string B[] = {"B1","B2","B3","B4","B5","T1","B6","B7","B8","B9","B10","T2","B11","B12","B13","B14","B15"}; 
// 35
string Node[] = {"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1","A10","A11","A12","A13","T2","A14","A15","A16","A17","A18",
				"B1","B2","B3","B4","B5","B6","B7","B8","B9","B10","B11","B12","B13","B14","B15"}; 

int matrix[N][N];

int dist[N][N];
int path[N][N];

int length=0;

string StrBegin,StrEnd;

void Floyd_Warshall();

int GetPos(string *array, string & str)
{
	int n=0;
	if (str[0] == 'B') {
		n+=20;
		array+=20;
	}
	if (str == "T1") return 9;
	if (str == "T2") return 14;

	while (*array != str) {
		array ++;
		n++;
	}
	return n;
}

void BuildGraph()
{
	// 初始化matrix
	for (int i=0; i 0)
			{
				dist[i][j] = matrix[i][j];
				path[i][j] = i;
			}
			else
			{
				dist[i][j] = INF;
				path[i][j] = -1;
			}
		}
	}
	// 2、Floyd核心三层循环
	for (int k=0; k dist[i][k] + dist[k][j])
				{
					dist[i][j] = dist[i][k] + dist[k][j];
					path[i][j] = path[k][j];
				}
			}
		}
	}
	// 3、输出结果
	int u = GetPos(Node,StrBegin);
	int v = GetPos(Node,StrEnd);
	while (u != v)
	{
		v = path[u][v];
		length++;
	}
	length ++; 
}


int main()
{

	//while(0)
	{
		length = 0;
		cin>>StrBegin>>StrEnd;
		if (StrBegin == StrEnd)
			cout<<"1"<