华为笔试题——地铁换乘

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
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

Mine >2h coding=>纯数学 佩服我自己。。。

#include<stdio.h>#include<string.h>#include<math.h>int main(){char a[20][4]={"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1","A10","A11","A12","A13","T2","A14","A15","A16","A17","A18"};char b[17][4]={"B1","B2","B3","B4","B5","T1","B6","B7","B8","B9",           "B10","T2","B11","B12","B13","B14","B15"};char c[4],d[4];int i,ca=100,cb=100,da=100,db=100,len;int len1,len2;// 比大小scanf("%s %s",c,d);for(i=0;i<20;i++){if(!strcmp(c,a[i])){ca=i;break;}}for(i=0;i<20;i++){if(!strcmp(d,a[i])){da=i;break;}}for(i=0;i<17;i++){if(!strcmp(c,b[i])){cb=i;break;}}for(i=0;i<17;i++){if(!strcmp(d,b[i])){db=i;break;}}//A-Aif(da!=100&&ca!=100){len=abs(da-ca)+1<10?(abs(da-ca)+1):(21-(abs(da-ca)));}//A-Bif(ca!=100&&db!=100){// 判断d 在B线的位置if(db<=4){len=(abs(9-ca)+1<10?(abs(9-ca)+1):(21-(abs(9-ca))))+5-db;}else if(db<=10){len1=(abs(9-ca)+1<10?(abs(9-ca)+1):(21-(abs(9-ca))))+db-5;len2=(abs(14-ca)+1<10?(abs(14-ca)+1):(21-(abs(14-ca))))+11-db;len =len1;if(len2<len)len=len2;}else{len=(abs(14-ca)+1<10?(abs(14-ca)+1):(21-(abs(14-ca))))+db-11;}}//B-Aif(cb!=100&&da!=100){if(cb>=12){len=cb-11+(abs(14-da)+1<10?(abs(14-da)+1):(21-(abs(14-da))));}else if (cb>=6){len1=11-cb+(abs(14-da)+1<10?(abs(14-da)+1):(21-(abs(14-da))));len2=cb-5+(abs(9-da)+1<10?(abs(9-da)+1):(21-(abs(9-da))));len=len1;if(len2<len)len=len2;}else {len1 =11-cb+(abs(14-da)+1<10?(abs(14-da)+1):(21-(abs(14-da))));len2 =5-cb+(abs(9-da)+1<10?(abs(9-da)+1):(21-(abs(9-da))));len=len1;if(len2<len1) len=len2;}}//B-Bif(cb!=100&&db!=100){   if((cb>=11&&db<=5)||(cb<=5&&db>=11)){len = abs(cb-db)+1-1;}elselen=abs(cb-db)+1;}printf("%d\n",len);return 0;}

运用科学的算法=>

#include<iostream>  #include<cstdio>  #include<cstring>  using namespace std;    //Floyed算法求任意两点之间的最短路径，算法复杂度O（n^3）虽然Floyed算法是求最短路径里面算法复杂度最大的算法，但写法简单，用于此处求任意两点之间的最短路合适  const int inf = 0x3f3f3f3f;//无穷大  C++struct Graph  {      char vertex[35][4];      int path[35][35];      int visited[35];  };  char s1[21][4]={"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1",      "A10","A11","A12","A13","T2","A14","A15","A16","A17","A18","A1"};  char s2[17][4]={"B1","B2","B3","B4","B5","T1","B6","B7","B8","B9",      "B10","T2","B11","B12","B13","B14","B15"};  char v[35][4]={"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"};    //关键在于如何建图  void CreateGraph(Graph * &G)  {      int i, j, k;      for (i = 0; i < 35; i++)      {          memcpy(G->vertex[i],v[i],sizeof(v[i]));  //v[i]-->G-vertex[i] 逐个拷贝        G->visited[i] = 0;                       //visit[i]置0    }      for (i = 0; i < 35; i++)                     //path[][]邻接矩阵 初始化大数inf    {          for (j = 0; j < 35; j++)          {              G->path[i][j] = inf;          }      }      for (k = 0; k < 20; k++)                     //相通则置为1 注意对称 A-A 互邻    {          for (i = 0;strcmp(s1[k],G->vertex[i])!=0; i++);          for (j = 0;strcmp(s1[k+1],G->vertex[j])!=0;j++);          G->path[i][j] = 1;          G->path[j][i] = 1;      }      for (k = 0; k < 16; k++)      {          for (i = 0;strcmp(s2[k],G->vertex[i])!=0; i++);   // B-B        for (j = 0; strcmp(s2[k+1],G->vertex[j])!=0; j++);          G->path[i][j] = 1;          G->path[j][i] = 1;      }  }  //Floyed算法求任意两点之间的最短路径  void Floyd(Graph * &G)  {      int i,j,k;      for (k = 0; k < 35; k++)      {          for (i = 0; i < 35; i++)          {              for (j = 0; j < 35; j++)              {                  if (G->path[i][k] + G->path[k][j] < G->path[i][j])                  {                      G->path[i][j] = G->path[i][k] + G->path[k][j];                  }              }          }      }  }    void ace(Graph *G)  {      char s1[4],s2[4];      int i,j;      cout<<"请输入起点站与终点站"<<endl;      cin>>s1>>s2;      for (i = 0;strcmp(s1,G->vertex[i])!=0;i++);      for (j = 0;strcmp(s2,G->vertex[j])!=0;j++);      cout<<G->path[i][j]+1<<endl;  }  int main()  {      Graph *G = new Graph;      CreateGraph(G);  //查看建的表 --deleteint m,n;for(m=0;m<35;m++){for(n=0;n<35;n++){if(G->path[m][n]==inf)G->path[m][n]=0;}}for(m=0;m<35;m++){for(n=0;n<35;n++){cout<<G->path[m][n];//printf("%d",G->path[m][n]);//printf("%d ",G->path[m][n]);}cout<<endl;}//Floyd(G);while(1)      {          ace(G);      }      system("pause");      return 0;  }