地铁换乘

</pre>题目简介： <br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">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）。 </span><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">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 </span><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">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 </span><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">输入：输入两个不同的站名 </span><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">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 </span><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">输入样例：A1 A3 </span><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><br style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px" /><p><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">输出样例：3 </span></p><p><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px"></span></p><p><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px"></span></p><p><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">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</span></p><p><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">所以搞了很久才搞出来</span></p><p><span style="color:rgb(17,17,17); font-family:Helvetica,Arial,sans-serif; font-size:13px; line-height:21.059999465942383px">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...</span></p><p><span style="font-family:Helvetica,Arial,sans-serif; font-size:12px; color:#111111"><span style="line-height:21.059999465942383px">我是用Dijkstra算法做的：</span></span></p><p><span style="font-family:Helvetica,Arial,sans-serif; font-size:12px; color:#111111"><span style="line-height:21.059999465942383px">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））</span></span></p><p><span style="font-family:Helvetica,Arial,sans-serif; font-size:12px; color:#111111"><span style="line-height:21.059999465942383px">然后用Dijkstra算法计算</span></span></p><p><span style="font-family:Helvetica,Arial,sans-serif; font-size:12px; color:#111111"><span style="line-height:21.059999465942383px"></span></span></p><p><span style="font-family:Helvetica,Arial,sans-serif; font-size:12px; color:#111111"><span style="line-height:21.059999465942383px"></span></span></p><pre name="code" class="cpp">#include<stdio.h>#include<string.h>#include<string>#include<iostream>using namespace std;typedef struct Tree{    //表示图的数据结构    int weight[36][36]; //权值    int mark[36];       //是否访问标记    int result[36];     //某站到所有站的最短距离（结果存储）}Tree;void init(Tree &train)  //生成相对应的图，初始化{    int i,j;    for (i=1; i <= 36; i++)        for (j=1; j<= 36; j++)            if (i == j)                train.weight[i][j] = 0;            else                train.weight[i][j] = 65535;    // A路线    for (i=1; i <= 8; i++){            train.weight[i][i+1] = 1;            train.weight[i+1][i] = 1;    }    for (i=10; i <= 12; i++){            train.weight[i][i+1] = 1;            train.weight[i+1][i] = 1;    }    for (i=14; i <= 17; i++){            train.weight[i][i+1] = 1;            train.weight[i+1][i] = 1;    }    train.weight[18][1] = 1;    train.weight[1][18] = 1;    // B路线    for (i=19; i <= 22; i++){            train.weight[i][i+1] = 1;            train.weight[i+1][i] = 1;    }    for (i=24; i <= 27; i++){            train.weight[i][i+1] = 1;            train.weight[i+1][i] = 1;    }    for (i=29; i <= 32; i++){            train.weight[i][i+1] = 1;            train.weight[i+1][i] = 1;    }    //Ｔ１　Ｔ２    train.weight[34][9] = 1;    train.weight[9][34] = 1;    train.weight[34][10] = 1;    train.weight[10][34] = 1;    train.weight[34][23] = 1;    train.weight[23][34] = 1;    train.weight[34][24] = 1;    train.weight[24][34] = 1;    train.weight[35][13] = 1;    train.weight[13][35] = 1;    train.weight[35][14] = 1;    train.weight[14][35] = 1;    train.weight[35][28] = 1;    train.weight[28][35] = 1;    train.weight[35][29] = 1;    train.weight[29][35] = 1;    for (i=1; i <= 36; i++)        train.mark[i] = 0;}//Dijkstra算法计算最短路径int foo(int a, int b, Tree train){    int i,j,mi,m,k;    for (i=1; i <= 36; i++){        train.result[i] = train.weight[a][i];        //printf("%d ",train.result[i]);    }    //printf("\n");    train.mark[a] = 1;    for (i=1; i <= 36; i++)        {            mi = 65535;            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）                if (!train.mark[j] && train.result[j] < mi)                {                    m = j;                    mi = train.result[j];                }            }            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                if (train.result[m] + train.weight[m][k] < train.result[k])                    train.result[k] = train.result[m] + train.weight[m][k]; //比之前的则更新                }            train.mark[m] = 1;        }    for (i=1; i <= 35; i++){                                                //结果输出        printf("%4d ",train.result[i]);    }    printf("\n");    return train.result[b]+1;}int main(){    string abt[] = {"A1", "A2", "A3", "A4", "A5", "A6", "A7", "A8", "A9", "A10", "A11", "A12", "A13", "A14", "A15", "A16", "A17", "A18",    "B1", "B2", "B3", "B4", "B5","B6", "B7", "B8", "B9", "B10", "B11", "B12", "B13", "B14", "B15", "T1", "T2"};    char s1[10],s2[10];    int a,b,i;    Tree train;    init(train);    scanf("%s%s",s1,s2);    for (i=0; i < 36; i++)    {        if (strcmp(s1,abt[i].c_str()) == 0)            a = i;        if (strcmp(s2,abt[i].c_str()) == 0)            b = i;    }    printf("%d %d\n",a+1,b+1);    printf("到相对应站的距离：\n");    for (i=1; i <= 35; i++){        printf("%4s ",abt[i-1].c_str());    }    printf("\n");    printf("\n最后结果：%d\n",foo(a+1,b+1,train));}