华为机试（8）

高级题样题：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 A10 A11 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 B10 T2 B11 B12 B13 B14 B15
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
输入样例：A1 A3
输出样例：3

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;
 int getRide(string pointA, string pointB)
 {
     string AA[20]={"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1","A10","A11","A12","A13","T2","A14","A15","A16","A17","A18"};
     vector<string> A(AA,AA+20);
     string BB[17]={"B1","B2","B3","B4","B5","T1","B6","B7","B8","B9","B10","T2","B11","B12","B13","B14","B15"};
     vector<string> B(BB,BB+17);

     int num1=1,num2 = 1;//顺向走向的站数num1，逆向走向的站数num2；
     int start;//起点的编号
     if((pointA[0]==‘A‘&&pointB[0]==‘A‘)||(pointA[0]==‘A‘&&pointB[0]==‘T‘)||(pointA[0]==‘T‘&&pointB[0]==‘A‘))//环形
     {
        int i = 0;
        while(A[i]!=pointA && i<20)
           i++;
        start = i;

        int j = start;//向前走的编号j
        while(A[j]!=pointB)
        {
           num1++;
           j++;
           if(j>=20)
               j -= 20;

        }

        int m = start;//向后走的编号j
        while(A[m]!=pointB)
        {
           num2++;
           m--;
           if(m<0)
               m += 20;

        }

        if(num1<num2)
            return num1;
        else
            return num2;

     }
     else if((pointA[0]==‘B‘&&pointB[0]==‘B‘)||(pointA[0]==‘T‘&&pointB[0]==‘B‘)||(pointA[0]==‘B‘&&pointB[0]==‘T‘))
     {
        num1 = 1;
        num2 = 1;
        int i = 0,a=0;
        int end;
        while(B[i]!=pointA && i<17)
           i++;
        start = i;

        while(B[i]!=pointB && i<17)
           a++;
        end = a;

        if(start>end)
        {
          int temp = start;
          start = end;
          end = temp;
        }

        if(start<=5 && end>=11)
            return end-start;
        else
            return end-start+1;

     }
     else if((pointA[0]==‘A‘&&pointB[0]==‘B‘)   || (pointA[0]==‘B‘&&pointB[0]==‘A‘))
     {
        if(pointA[0]==‘B‘)
        {
          string temp = pointA;
          pointA = pointB;
          pointB = temp;
        }
        int i = 0;
        while(B[i]!=pointB && i<17)
           i++;
        start = i;
        //B分别遇到T1、T2后选择A线路顺着或者逆着走，共有4个答案；
        vector<int> res(4,0); // 分别记录T1顺时针、T1逆时针、T2顺时针、T2逆时针
        if(i<5)//到T1、T2的初始值的计算
        {
         res[0] = 5 - start+1;
         res[1]=res[0];
         res[2] = 11 - start+1;
         res[3]=res[2];

        }
        else if(i>5 && i<11)
        {
           res[0] = start-5+1;
           res[1]=res[0];
           res[2] = 11 - start+1;
           res[3]=res[2];
        }
        else
        {
           res[0] = start-5+1;
           res[1]=res[0];
           res[2] = start-11+1;
           res[3]=res[2];
        }//初始值计算完毕

        //T1到A1的顺时针距离
         int j0 = 9;//向前走的编号j0
        while(A[j0]!=pointA)
        {
           res[0]++;
           j0++;
           if(j0>=20)
               j0 -= 20;
        }
        //T1到A1的逆时针距离
        int j1 = 9;
        while(A[j1]!=pointA)
        {
           res[1]++;
           j1--;
           if(j1<0)
               j1 += 20;
        }
        //T2到A1的逆时针距离
         int j2 = 14;//向前走的编号j0
        while(A[j2]!=pointA)
        {
           res[2]++;
           j2++;
           if(j2>=20)
               j2 -= 20;
        }
        //T2到A1的顺时针距离
        int j3 = 14;
        while(A[j3]!=pointA)
        {
           res[3]++;
           j3--;
           if(j3<0)
               j3 += 20;
        }

        sort(res.begin(),res.end());
        return res[0];
     }
     else if((pointA=="T1"&&pointB=="T2") ||(pointA=="T2"&&pointB=="T1"))
         return 6;
 }
void main()
{
    string A="T2";
    string B="T1";
    int n = getRide(A,B);
    cout<<n<<endl;
}

要把各种情况都考虑到，开始写时把框架先写好，然后往框架中填代码，写一部分测试一部分，如果有拷贝的代码，写时一定要仔细检查确保变量正确无误，否则浪费时间调试。

思路清晰，情况考虑全，就这两点！

华为机试（8）