c语言课程找代写有偿
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)。输入的站点名必须与线路站点名一致，不一致时提示用户重新输入
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define NODENUM 50
#define MAXWEIGHT 65535
typedef struct Node {
int flag;
int weight;
} Matrix[NODENUM][NODENUM];
typedef struct MGraph {
int arcnum;
int vertexnum;
Matrix adjmatrix;
} MGraph, *pMGraph;
typedef enum {
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
} index;
char* str[] = {"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"};
void CreateSubRailWay(pMGraph mg) {
int n = sizeof(str) / sizeof(char*);
mg->vertexnum = n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
mg->adjmatrix[i][j].weight = MAXWEIGHT;
}
}
for (int i = A1; i <= A17; i++) {
mg->adjmatrix[i][(i + 1)].weight = 1;
mg->adjmatrix[i + 1][i].weight = 1;
}
mg->adjmatrix[A18][A1].weight = 1;
mg->adjmatrix[A1][A18].weight = 1;
for (int i = B1; i <= B4; i++) {
mg->adjmatrix[i][i + 1].weight = 1;
mg->adjmatrix[i + 1][i].weight = 1;
}
mg->adjmatrix[B5][T1].weight = 1;
mg->adjmatrix[T1][B5].weight = 1;
mg->adjmatrix[T1][B6].weight = 1;
mg->adjmatrix[B6][T1].weight = 1;
for (int i = B6; i <= B9; i++) {
mg->adjmatrix[i][i + 1].weight = 1;
mg->adjmatrix[i + 1][i].weight = 1;
}
mg->adjmatrix[B10][T2].weight = 1;
mg->adjmatrix[T2][B10].weight = 1;
mg->adjmatrix[T2][B11].weight = 1;
mg->adjmatrix[B11][T2].weight = 1;
for (int i = B11; i <= B14; i++) {
mg->adjmatrix[i][i + 1].weight = 1;
mg->adjmatrix[i + 1][i].weight = 1;
}
}
int Path[NODENUM];
int ShortestPath[NODENUM];
int used_dij[NODENUM];
int Dijk(pMGraph mg, int startpos) {
int n = mg->vertexnum;
for (int i = 0; i < n; i++) {
Path[i] = startpos;
ShortestPath[i] = mg->adjmatrix[startpos][i].weight;
used_dij[i] = 0;
}
ShortestPath[startpos] = 0;
used_dij[startpos] = 1;
Path[startpos] = startpos;
for (int i = 1; i < n; i++) {
int min = MAXWEIGHT;
int k;
for (int j = 0; j < n; j++) {
if (used_dij[j] != 1 && min > ShortestPath[j]) {
min = ShortestPath[j];
k = j;
}
}
used_dij[k] = 1;
for (int j = 0; j < n; j++) {
if (used_dij[j] == 0 &&
mg->adjmatrix[k][j].weight + min < ShortestPath[j]) {
ShortestPath[j] = min + mg->adjmatrix[k][j].weight;
Path[j] = k;
}
}
}
return 0;
}
int main(int argc, char* argv[]) {
pMGraph mg = (pMGraph)malloc(sizeof(MGraph));
CreateSubRailWay(mg);
printf("输入:\n");
char start[10], end[10];
scanf("%s", start);
scanf("%s", end);
int startpos, endpos;
for (int i = 0; i < sizeof(str) / sizeof(char*); i++) {
if (strcmp(start, str[i]) == 0) {
startpos = i;
}
if (strcmp(end, str[i]) == 0) {
endpos = i;
}
}
Dijk(mg, startpos);
printf("站数为%d\n", ShortestPath[endpos] + 1);
int len = sizeof(str) / sizeof(char*);
int sitecount = 0;
int sites[sizeof(str) / sizeof(char*)];
int now = endpos;
while (now != startpos) {
sites[sitecount++] = now;
now = Path[now];
}
sites[sitecount] = now;
printf("路线为");
for (int i = 0; i <= sitecount; i++) {
if (i <= sitecount - 1) {
printf("%s->", str[sites[sitecount - i]]);
}else{
printf("%s\n", str[sites[sitecount - i]]);
}
}
return 0;
}
运行结果：
- 你可以看下这个问题的回答https://ask.csdn.net/questions/7614113
- 我还给你找了一篇非常好的博客，你可以看看是否有帮助，链接：c语言：输入两个整形数并打印，如果用户不慎输入了非法字符，那么程序提示“输入数据类型错误”。
- 除此之外, 这篇博客: ＜栈和队列＞题目练习《数据结构(C语言版)》中的 5.用无头单链表存储队列，其头指针指向队头结点，尾指针指向队尾结点，则在进行出队操作时（ ） 部分也许能够解决你的问题, 你可以仔细阅读以下内容或跳转源博客中阅读:
A.仅修改队头指针
B.队头、队尾指针都要修改
C.队头、队尾指针都可能要修改
D.仅修改队尾指针
- 您还可以看一下 钟洪发老师的鸿蒙设备开发之C知识点全面梳理课程中的 实现插入文件内容小节, 巩固相关知识点
- 以下回答由chatgpt基于相关博客总结生成:
我可以帮你完成这个任务。首先，你需要构建一个地铁网络的数据结构来表示所有的站点和线路，并且需要实现一个函数来计算最短路径。下面是一个可能的解决方案：
- 定义地铁网络的数据结构：
#define MAX_STATIONS 100 typedef struct { char name[50]; // 站点名称 } Station; typedef struct { int start; // 起始站点索引 int end; // 终点站点索引 int stations[MAX_STATIONS]; // 地铁经过的站点索引 int count; // 经过的站点数量 } Path; typedef struct { Station stations[MAX_STATIONS]; // 所有站点 int numStations; // 站点数量 int adjacencyMatrix[MAX_STATIONS][MAX_STATIONS]; // 邻接矩阵表示线路 } SubwayNetwork;- 实现函数来初始化地铁网络：
void initSubwayNetwork(SubwayNetwork* network) { // 初始化站点数量为0 network->numStations = 0; // 初始化邻接矩阵 for (int i = 0; i < MAX_STATIONS; i++) { for (int j = 0; j < MAX_STATIONS; j++) { network->adjacencyMatrix[i][j] = 0; } } } void addStation(SubwayNetwork* network, const char* name) { // 检查是否已经达到最大站点数量 if (network->numStations >= MAX_STATIONS) { printf("站点数量达到最大限制\n"); return; } // 添加新的站点 strcpy(network->stations[network->numStations].name, name); network->numStations++; } void addConnection(SubwayNetwork* network, int station1, int station2) { network->adjacencyMatrix[station1][station2] = 1; network->adjacencyMatrix[station2][station1] = 1; }- 实现函数来计算最短路径：
void findShortestPath(SubwayNetwork* network, const char* startStation, const char* endStation, Path* shortestPath) { int start = -1; int end = -1; // 查找起点和终点的索引 for (int i = 0; i < network->numStations; i++) { if (strcmp(network->stations[i].name, startStation) == 0) { start = i; } if (strcmp(network->stations[i].name, endStation) == 0) { end = i; } } // 如果没找到站点，提醒用户重新输入 if (start == -1 || end == -1) { printf("输入的站点名与线路中的站点名不一致\n"); return; } // 使用广度优先搜索算法找到最短路径 int visited[MAX_STATIONS] = {0}; // 标记是否已经访问过 int queue[MAX_STATIONS] = {0}; // 存储待访问的站点 int prev[MAX_STATIONS] = {-1}; // 记录访问路径 int rear = 0; // 队列尾 int front = 0; // 队列头 int found = 0; // 是否找到终点 visited[start] = 1; queue[rear++] = start; while (front < rear) { int current = queue[front++]; if (current == end) { found = 1; // 找到终点 break; } for (int i = 0; i < network->numStations; i++) { if (network->adjacencyMatrix[current][i] == 1 && visited[i] == 0) { visited[i] = 1; queue[rear++] = i; prev[i] = current; } } } // 根据访问路径生成最短路径 shortestPath->count = 0; int current = end; while (current != -1) { shortestPath->stations[shortestPath->count++] = current; // 存储路径上的站点 current = prev[current]; } shortestPath->start = start; shortestPath->end = end; } void printPath(SubwayNetwork* network, const Path* path) { printf("从 %s 到 %s 的最短路径（包括换乘站点）：\n", network->stations[path->start].name, network->stations[path->end].name); for (int i = path->count - 1; i >= 0; i--) { printf("%s ", network->stations[path->stations[i]].name); } printf("\n"); }- 使用上述代码来解决你的问题：
int main() { // 初始化地铁网络 SubwayNetwork network; initSubwayNetwork(&network); // 添加站点和线路 addStation(&network, "站点1"); addStation(&network, "站点2"); addStation(&network, "站点3"); addConnection(&network, 0, 1); addConnection(&network, 1, 2); // 输入起点和终点 char startStation[50]; char endStation[50]; printf("请输入起始站点名："); scanf("%s", startStation); printf("请输入终点站点名："); scanf("%s", endStation); // 计算最短路径 Path shortestPath; findShortestPath(&network, startStation, endStation, &shortestPath); // 打印最短路径 printPath(&network, &shortestPath); return 0; }这样就完成了一个简单的地铁最短路径计算程序。请注意，上述代码只是一种解决方案，实际上还有很多其他方法来解决这个问题。如果有任何问题，请随时向我提问。
简单写了下，你参考参考
上图
代码
#include <iostream>
#include <string>
#include <vector>
using namespace std;
// 定义地铁线路
vector<string> line1 = {
"临川路",
"北商报站",
"书院门站",
"观音阁站",
"东大街站",
"钟鼓楼站",
"崇安门站",
"东市口站",
"大观园站",
"江宁路站",
"常熟路站",
"五塘站",
};
vector<string> line2 = {
"金陵大厦",
"人民广场",
"建邺学院站",
"大曹口站",
"邗江站",
"鼓楼站",
"丰泽站",
"振华站",
"钟南山站",
"珠江路站",
"华夏高新园区",
"甪直站"
};
int main() {
string start, end;
cout << "请输入出发站点:";
cin >> start;
cout << "请输入目的地站点:";
cin >> end;
// 检查站点名称是否正确
bool valid_start = false, valid_end = false;
for (string name : line1) {
if (name == start) valid_start = true;
if (name == end) valid_end = true;
}
for (string name : line2) {
if (name == start) valid_start = true;
if (name == end) valid_end = true;
}
// 错误则重新输入
while(!valid_start || !valid_end) {
if(!valid_start) {
cout << "出发站点输入错误,请重新输入" << endl;
cin >> start;
valid_start = false;
}
if (!valid_end){
cout << "目的地站点输入错误,请重新输入:" << endl;
cin >> end;
valid_end = false;
}
// 重新检查站点名称是否存在
for (string name : line1) {
if (name == start) valid_start = true;
if (name == end) valid_end = true;
}
for (string name : line2) {
if (name == start) valid_start = true;
if (name == end) valid_end = true;
}
}
// 查找最少站数
int minStations = INT_MAX;
if (line1.size() > 0 && line2.size() > 0) {
int startIndex = -1;
for (int i = 0; i < line1.size(); i++) {
if (line1[i] == start) {
startIndex = i;
break;
}
}
int endIndex = -1;
for (int i = 0; i < line1.size(); i++) {
if (line1[i] == end) {
endIndex = i;
break;
}
}
if (startIndex != -1 && endIndex != -1) {
int stations = abs(startIndex - endIndex);
minStations = min(stations, minStations);
}
}
if (line2.size() > 0) {
int startIndex = -1;
for (int i = 0; i < line2.size(); i++) {
if (line2[i] == start) {
startIndex = i;
break;
}
}
int endIndex = -1;
for (int i = 0; i < line2.size(); i++) {
if (line2[i] == end) {
endIndex = i;
break;
}
}
if (startIndex != -1 && endIndex != -1) {
int stations = abs(startIndex - endIndex);
minStations = min(stations, minStations);
}
}
cout << "最少需要" << minStations << "站" << endl;
}
这就是一个最短路径的问题，结合具体的例子使用程序实现而已。在具体的实现思路方面，根据题目中你的图，使用图的最短路径广度优先算法即可实现。
具体的实现代码可以参考：
地铁最短线路(代码实现)：https://www.cnblogs.com/DarkWalkerLynn/p/13930526.html
https://github.com/wengzheng/subway
可以参考我的博客例子