题目简介：
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
这种鬼题刚开始看的时候没什么头绪，看上去就像是要用数据结构来做。不过个人数据结构又不是很好
所以搞了很久才搞出来
这题有挺多方法解决的：可以穷举，可以用Dijkstra算法，可以用Floyd算法...
我是用Dijkstra算法做的：
首先生成相对应的图：用矩阵表示（也可以用链表（可能相对复杂些））
然后用Dijkstra算法计算
1 2 3 #include<stdio.h> 4 #include<string.h> 5 #include<string> 6 #include<iostream> 7 using namespace std; 8 typedef struct Tree{ //表示图的数据结构 9 int weight[36][36]; //权值 10 int mark[36]; //是否访问标记 11 int result[36]; //某站到所有站的最短距离（结果存储） 12 }Tree; 13 14 void init(Tree &train) //生成相对应的图，初始化 15 { 16 int i,j; 17 for (i=1; i <= 36; i++) 18 for (j=1; j<= 36; j++) 19 if (i == j) 20 train.weight[i][j] = 0; 21 else 22 train.weight[i][j] = 65535; 23 // A路线 24 for (i=1; i <= 8; i++){ 25 train.weight[i][i+1] = 1; 26 train.weight[i+1][i] = 1; 27 } 28 for (i=10; i <= 12; i++){ 29 train.weight[i][i+1] = 1; 30 train.weight[i+1][i] = 1; 31 } 32 for (i=14; i <= 17; i++){ 33 train.weight[i][i+1] = 1; 34 train.weight[i+1][i] = 1; 35 } 36 train.weight[18][1] = 1; 37 train.weight[1][18] = 1; 38 // B路线 39 for (i=19; i <= 22; i++){ 40 train.weight[i][i+1] = 1; 41 train.weight[i+1][i] = 1; 42 } 43 for (i=24; i <= 27; i++){ 44 train.weight[i][i+1] = 1; 45 train.weight[i+1][i] = 1; 46 } 47 for (i=29; i <= 32; i++){ 48 train.weight[i][i+1] = 1; 49 train.weight[i+1][i] = 1; 50 } 51 //Ｔ１ Ｔ２ 52 train.weight[34][9] = 1; 53 train.weight[9][34] = 1; 54 train.weight[34][10] = 1; 55 train.weight[10][34] = 1; 56 train.weight[34][23] = 1; 57 train.weight[23][34] = 1; 58 train.weight[34][24] = 1; 59 train.weight[24][34] = 1; 60 train.weight[35][13] = 1; 61 train.weight[13][35] = 1; 62 train.weight[35][14] = 1; 63 train.weight[14][35] = 1; 64 train.weight[35][28] = 1; 65 train.weight[28][35] = 1; 66 train.weight[35][29] = 1; 67 train.weight[29][35] = 1; 68 for (i=1; i <= 36; i++) 69 train.mark[i] = 0; 70 } 71 72 //Dijkstra算法计算最短路径 73 int foo(int a, int b, Tree train) 74 { 75 int i,j,mi,m,k; 76 for (i=1; i <= 36; i++){ 77 train.result[i] = train.weight[a][i]; 78 //printf("%d ",train.result[i]); 79 } 80 //printf(" "); 81 train.mark[a] = 1; 82 for (i=1; i <= 36; i++) 83 { 84 mi = 65535; 85 for (j=1; j <= 36; j++){ //搜寻当前未访问的最短路径(作为下一访问点） 86 if (!train.mark[j] && train.result[j] < mi) 87 { 88 m = j; 89 mi = train.result[j]; 90 } 91 } 92 for (k=1; k <= 36; k++){ //比较（经过当前访问点的距离）与之前最短路径的距离 93 if (train.result[m] + train.weight[m][k] < train.result[k]) 94 train.result[k] = train.result[m] + train.weight[m][k]; //比之前的则更新 95 } 96 train.mark[m] = 1; 97 98 } 99 for (i=1; i <= 35; i++){ //结果输出 100 printf("%4d ",train.result[i]); 101 } 102 printf(" "); 103 return train.result[b]+1; 104 } 105 106 int main() 107 { 108 string abt[] = {"A1", "A2", "A3", "A4", "A5", "A6", "A7", "A8", "A9", "A10", "A11", "A12", "A13", "A14", "A15", "A16", "A17", "A18", 109 "B1", "B2", "B3", "B4", "B5","B6", "B7", "B8", "B9", "B10", "B11", "B12", "B13", "B14", "B15", "T1", "T2"}; 110 char s1[10],s2[10]; 111 int a,b,i; 112 Tree train; 113 init(train); 114 scanf("%s%s",s1,s2); 115 for (i=0; i < 36; i++) 116 { 117 if (strcmp(s1,abt[i].c_str()) == 0) 118 a = i; 119 if (strcmp(s2,abt[i].c_str()) == 0) 120 b = i; 121 } 122 printf("%d %d ",a+1,b+1); 123 printf("到相对应站的距离： "); 124 for (i=1; i <= 35; i++){ 125 printf("%4s ",abt[i-1].c_str()); 126 } 127 printf(" "); 128 printf(" 最后结果：%d ",foo(a+1,b+1,train)); 129 130 } 131