Data Structure Course Design (C Language)-Qingdao Agricultural University Campus Tour Guide System
Including the complete course design process and source code
1 Problem description
Use an undirected network to represent the plane map of your school’s campus attractions (about 10). The vertices in the figure represent the main scenic spots, and store the number, name, introduction and other information of the scenic spots. information. It is required to be able to answer questions about scenic spot introductions, tour routes, etc.
Requirements:
1 Inquire about relevant information of each scenic spot.
2 Query the shortest path between any two scenic spots in the graph.
3 Query all paths between any two scenic spots in the graph.
4 Add, delete, and update information about scenic spots and roads.
2 problem analysis
The campus navigation system is a system designed based on the relevant knowledge of the graph in the data structure. Taking the campus attractions as the nodes of the graph, the paths between the vertices as the edges of the graph, and the distance between the edges as the weights between the nodes, it is abstracted into an undirected weighted graph. Designed to quickly find the shortest path and shortest distance from one attraction to another, to help people find various school attractions more conveniently and facilitate visits.
Figure 2-1 Map of Qingdao Agricultural University
3 design ideas
Figure 3–1 Functional Diagram
3.1 Two core algorithms
Dijkstra's algorithm: Let G=(V, E) be a weighted directed graph, divide the vertex set V in the graph into two groups, and the first group is the set of vertices whose shortest path has been obtained (indicated by S, initially There is only one source point in S, and every time a shortest path v,...,u is obtained, u will be added to the set S until all vertices are added to S, and the algorithm is over), the second group is the remaining Determine the set of vertices (denoted by U) of the shortest path, and add the second group of vertices to S in the ascending order of the length of the shortest path.
Floyd's Algorithm: Dynamic programming algorithm. Find the shortest path from point i to point j. There are two possibilities for the shortest path from i to j, one is directly from i to j, and the other is from i to j through several nodes k. Suppose Dis(i,j) is the shortest path distance from node u to node v, for each node k, check whether Dis(i,k) + Dis(k,j) < Dis(i,j) holds true, if Established, prove that the path from i to k and then to j is shorter than the path from i to j directly, we set Dis(i,j) = Dis(i,k) + Dis(k,j), when all nodes are traversed k, Dis(i,j) records the distance of the shortest path from i to j.
3.2 Structure and variables
//青岛农业大学校园管理系统
//2021/10/6
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define Infinity 65535 //表示无穷大
#define MaxNumber 23 //用于邻接矩阵
#define vertex 15 //顶点个数
typedef struct side//边的权值
{
int wet;//权值
}side,wetmatrix[MaxNumber][MaxNumber];//边的邻接矩阵类型
typedef struct vetinf//顶点信息
{
int number;//顶点编号
char name[64];//顶点名称
char intro[256];//顶点介绍
}vetinf;
typedef struct mapstr//图结构信息
{
vetinf vets[MaxNumber];//顶点数组
wetmatrix mat;//邻接矩阵
int vetnum,sidenum;
}mapstr;
//全局变量
mapstr campus;//图结构变量（学校校园）
int d[30];
int visited[50];
int shortest[MaxNumber][MaxNumber];//定义全局变量存储最小路径
int pathh[MaxNumber][MaxNumber];//定义存储路径
3.3 Main working functions
//1.图的初始化
mapstr initmap()
//2.查询景点在图中的序号
int locatevet(mapstr m,int v)
//3.查询输入序号l,n间的长度不超过10个景点的路径
void path(mapstr m,int l,int n,int k)
//4.查询两景点的所有路径
int allpath(mapstr m)
//5.迪杰斯特拉算法求单源最短路径
void shortestpath(mapstr m)
//6.主页
void menu()
//7.重新构造图
int creatmap(mapstr *m)
//8.更改图部分信息
int newmap(mapstr *m)
//9.增加一条边
int incside(mapstr *m)
//10.增加一个结点
int incvet(mapstr *m)
//11.删除一个结点
int delvet(mapstr *m)
//12.删除一条边
int delside(mapstr *m)
//13.输出图的邻接矩阵的值
void printmapstr(mapstr m)
//14.图的操作主函数
int changemap(mapstr *m)
//15.用户登录
int userlog()
//16.弗洛伊德算法
void floyd(mapstr m)
//17.输出数组
void printarray()
//18.输出最短路径
void display(mapstr m,int i,int j)
//19任意两点间距离（6-19）
int shortdistance(mapstr m)
//20显示所有点信息
void compusinfor(mapstr m)
//21青岛农业大学地图
void schoolmap()
//22用户界面
void mainwork()
//22主函数
int main()
4 complete program source code
//青岛农业大学校园管理系统
//完整源代码
//2021/10/6
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define Infinity 65535 //表示无穷大
#define MaxNumber 23 //用于邻接矩阵
#define vertex 15 //顶点个数
typedef struct side//边的权值
{
int wet;//权值
}side,wetmatrix[MaxNumber][MaxNumber];//边的邻接矩阵类型
typedef struct vetinf//顶点信息
{
int number;//顶点编号
char name[64];//顶点名称
char intro[256];//顶点介绍
}vetinf;
typedef struct mapstr//图结构信息
{
vetinf vets[MaxNumber];//顶点数组
wetmatrix mat;//邻接矩阵
int vetnum,sidenum;
}mapstr;
//全局变量
mapstr campus;//图结构变量（学校校园）
int d[30];
int visited[50];
int shortest[MaxNumber][MaxNumber];//定义全局变量存储最小路径
int pathh[MaxNumber][MaxNumber];//定义存储路径
//1.图的初始化
mapstr initmap()
{
mapstr m;//构件图m
int i=0,j=0;
m.vetnum=15;//定义顶点个数
m.sidenum=23;//定义边的条数
for (i=1;i<=vertex;i++)//依次设置顶点信息
m.vets[i].number=i;
//输入顶点信息
strcpy(m.vets[1].name,"西苑宿舍");strcpy(m.vets[1].intro,"位于学校最西端，学生住宿场所");
strcpy(m.vets[2].name,"西苑餐厅");strcpy(m.vets[2].intro,"为西苑学生提供就餐场所");
strcpy(m.vets[3].name,"地下通道");strcpy(m.vets[3].intro,"连接东苑与西苑，上方为公路");
strcpy(m.vets[4].name,"文体馆");strcpy(m.vets[4].intro,"学校举办文体活动的场所");
strcpy(m.vets[5].name,"体育场");strcpy(m.vets[5].intro,"五环体育场");
strcpy(m.vets[6].name,"东苑宿舍");strcpy(m.vets[6].intro,"东苑学生住宿场所");
strcpy(m.vets[7].name,"东苑餐厅");strcpy(m.vets[7].intro,"有博爱海都润兴三大餐厅，为东苑学生提供就餐场所");
strcpy(m.vets[8].name,"农大花园");strcpy(m.vets[8].intro,"在这里可以欣赏到学校的一处优美的景色");
strcpy(m.vets[9].name,"虹子湖");strcpy(m.vets[9].intro,"学校的湖，景色优美");
strcpy(m.vets[10].name,"足球场");strcpy(m.vets[10].intro,"位于学校最东端，为学生踢球场所");
strcpy(m.vets[11].name,"科技楼信息楼化学楼");strcpy(m.vets[11].intro,"这里有学校的信息楼,科技楼,化学楼");
strcpy(m.vets[12].name,"办公楼图书馆生物楼");strcpy(m.vets[12].intro,"这里有学校的办公楼,图书馆,生物楼");
strcpy(m.vets[13].name,"教学楼");strcpy(m.vets[13].intro,"学生学习的场所");
strcpy(m.vets[14].name,"工程楼,文经楼");strcpy(m.vets[14].intro,"这里有学校的工程楼,文经楼");
strcpy(m.vets[15].name,"虹子广场");strcpy(m.vets[15].intro,"刚步入学校南门的广场，举办各种活动的地方");
for (i=1;i<=vertex;i++)
for (j=1;j<=vertex;j++)
m.mat[i][j].wet=Infinity;//初始化图的邻接矩阵
m.mat[1][2].wet=10;m.mat[2][3].wet=40;m.mat[3][4].wet=20;m.mat[3][8].wet=40;m.mat[3][11].wet=40;
m.mat[4][5].wet=20;m.mat[4][8].wet=20;m.mat[5][6].wet=30;m.mat[5][8].wet=30;m.mat[6][7].wet=10;
m.mat[6][9].wet=70;m.mat[7][9].wet=60;m.mat[7][10].wet=40;m.mat[8][9].wet=10;m.mat[9][10].wet=30;
m.mat[9][13].wet=15;m.mat[9][14].wet=30;m.mat[10][14].wet=10;m.mat[11][12].wet=50;m.mat[11][15].wet=70;
m.mat[12][13].wet=5;m.mat[12][15].wet=10;m.mat[13][14].wet=40;
for(i=1;i<=vertex;i++)//无向带权图是对称矩阵,给其另一半赋值
for(j=1;j<=vertex;j++)
m.mat[j][i].wet=m.mat[i][j].wet;
return m;
}
//2.查询景点在图中的序号
int locatevet(mapstr m,int v)
{
int i;
for(i=0;i<=m.vetnum;i++)
if(v==m.vets[i].number) return i;//找到返回顶点i
return -1;//未找到
}
//3.查询输入序号l,n间的长度不超过10个景点的路径
void path(mapstr m,int l,int n,int k)
{
int s,t=k+1;int length=0;//t用于存储路径上下一顶点对应的d[]数组元素的下标
if(d[k]==n&&k<8)//若d[k]是终点且景点个数<8，则输出该路径
{
for(s=0;s<k;s++)
{
length=length+m.mat[d[s]][d[s+1]].wet;
}
if(length<200)//打印路径小于200（定长）的路径
{
for(s=0;s<k;s++)//输出该路径，s=0时为起点m
{
printf("%d%s--->",d[s],m.vets[d[s]].name);
}
printf("%d%s ",d[s],m.vets[d[s]].name);//输出最后一个顶点
printf("总路线长为%d米\n\n",length);
}
}
else
{
s=1;
while(s<=m.vetnum)//从第m个顶点，访问所有顶点是否有路径
{
if((m.mat[d[k]][s].wet<Infinity)&&(visited[s]==0))//顶点有边且未被访问
{
visited[s]=1;
d[k+1]=s;//存储顶点编号
path(m,l,n,t);
visited[s]=0;//将找到的路径上的顶点的访问标志重新设置为，便于探究新的路径
}
s++;//试验下一顶点s开始是否有到终点的路径；
}
}
}
//4.查询两景点的所有路径
int allpath(mapstr m)
{
int k,i,j,l,n;
printf("\n\n请输入您想要查询的两个景点的编号：\n\n");
scanf("%d%d",&i,&j);printf("\n\n");
l=locatevet(m,i);//locatevet 确定该顶点是否存在。若存在，返回该顶点编号。
n=locatevet(m,j);
d[0]=l;//路径起点l(字母).(d[]数组为全局变量)
for(k=0;k<vertex;k++)
visited[k]=0;
visited[l]=1;
path(m,l,n,0);
return 1;
}
//5.迪杰斯特拉算法求单源最短路径
void shortestpath(mapstr m)
{
int v0,v,w,k=1,min,t,p;
int final[MaxNumber];//final[w]=1表示已经求得顶点V0到Vw的的最短路径
int Pathside[MaxNumber];//用于存储最短路径下标的数组
int ShortPathwet[MaxNumber];//用于存储到各点最短路径的权值和
printf("\n请输入起始景点的编号：");
scanf("%d",&v0);
printf("\n\n");
while(v0<0||v0>vertex)//判断是否输入正确
{
printf("\n您输入的景点编号不存在\n");
printf("请重新输入：");
scanf("%d",&v0);
}
for(v=1;v<=m.vetnum;v++)//数组初始化
{
final[v]=0;//全部顶点初始化为未找到路径
ShortPathwet[v]=m.mat[v0][v].wet;//将与v0有连线的路径加上权值
Pathside[v]=0;//初始化路径数组为0
}
ShortPathwet[v0]=0;
final[v0]=1;
//Dijkstr算法主体
for(v=1;v<=m.vetnum;v++)
{
min=Infinity;
for(w=1;w<=m.vetnum;w++)//找出离当前指向顶点最近的点
{
if(!final[w]&&ShortPathwet[w]<min)//未被访问且存在边
{
k=w;
min=ShortPathwet[w];
}
}
final[k]=1;//将找到的离当前顶点最近的置1
//修正
for(w=1;w<=m.vetnum;w++)
{
if(!final[w]&&(min+m.mat[k][w].wet<ShortPathwet[w]))
{
ShortPathwet[w]=min+m.mat[k][w].wet;//修改当前最优路径长度
Pathside[w]=k;//存放前驱结点
}
}
}
printf("打印P数组:"); //打印p数组
for(t=1;t<=m.vetnum;t++)
{
printf("%d ",Pathside[t]);
}
printf("\n\n");
printf("打印S数组:"); //打印s数组
for(t=1;t<=m.vetnum;t++)
{
printf("%d ",ShortPathwet[t]);
}
printf("\n\n");
//打印最短路径
for(t=1;t<=m.vetnum;t++)
{
p=t;
if(t!=v0)
{
printf("%d%s",t,m.vets[t].name);
for(w=1;w<=m.vetnum;w++)
{
if(Pathside[p]!=0)
{
printf("<--%d%s",Pathside[p],m.vets[p].name);
p=Pathside[p];
}
}
printf("<--%d%s",v0,m.vets[v0].name);
printf("\n总路线长为%d米\n\n",ShortPathwet[t]);
}
}
}
//6.主页
void menu()
{
printf(" ┌──────────────────────────────────────────────────────┐\n");
printf(" │ ╭ ═══════════════════════════════════════════════ ╮ │\n");
printf(" │ ││ 欢 迎 使 用 校 园 导 游 系 统 ││ │\n");
printf(" │ ╰ ═══════════════════════════════════════════════ ╯ │\n");
printf(" │ 欢迎来到 │\n");
printf(" │ 青岛农业大学 │\n");
printf(" │ 菜 单 选 择 │\n");
printf(" │ *************************************************** │\n");
printf(" │ * 1.主页 ** 2.查看游览路线 * │\n");
printf(" │ *************************************************** │\n");
printf(" │ * 3.查询景点间最短路径 ** 4.查询景点间所有路径* │\n");
printf(" │ *************************************************** │\n");
printf(" │ * 5.学校景点介绍 ** 6.学校地图 * │\n");
printf(" │ *************************************************** │\n");
printf(" │ * 7.更改图信息 ** 0.退出 * │\n");
printf(" │ *************************************************** │\n");
printf(" └──────────────────────────────────────────────────────┘\n");
}
//以下是修改图的相关信息。
//7.重新构造图
int creatmap(mapstr *m)
{
int i,j,l,n,v0,v1,distance;
printf("请输入图的顶点数和边数:\n");
scanf("%d %d",&m->vetnum,&m->sidenum);
printf("请输入顶点信息: \n");
for(i=0;i<=m->vetnum;i++)//输入各顶点对应的景点信息
{
printf("请输入景点编号:");
scanf("%d",&m->vets[i].number);
printf("请输入景点名称:");
scanf("%s",&m->vets[i].name);
scanf("请输入景点简介:");
scanf("%s",&m->vets[i].intro);
}
for(i=0;i<=m->vetnum;i++)//权值初始化为无穷
for(j=0;j<=m->vetnum;j++)
m->mat[i][j].wet=Infinity;
printf("请输入图中各景点边的信息 \n");
for(i=1;i<=m->sidenum;i++)
{
printf("\n请输入第%d条边的起点，终点，长度为：",i);
scanf("%d %d %d",&v0,&v1,&distance);
l=locatevet(campus,v0);
n=locatevet(campus,v1);
if(l>0&&n>0)
{
m->mat[l][n].wet=distance;
m->mat[n][l].wet=m->mat[l][n].wet;
}
}
return 1;
}
//8.更改图部分信息
int newmap(mapstr *m)
{
int changenum,i,l,n,t,distance,v0,v1;
printf("\n下面请输入你要修改的景点的个数：\n");//修改定点对应的景点
scanf("%d",&changenum);
while(changenum<0||changenum>vertex)
{
printf("\n你的输入有误，请重新输入");
scanf("%d",&changenum);
}
for(i=0;i<changenum;i++)
{
printf("\n请输入景点编号：");
scanf("&d",&l);
t=locatevet(campus,l);
printf("\n请输入修改后景点的名称：");
scanf("%s",&m->vets[t].name);
printf("\n请输入修改后景点的简介:");
scanf("%s",&m->vets[t].intro);
}
printf("\n下面请输入你要修改的边的个数");
scanf("%d",&changenum);
while(changenum<0||changenum>vertex);
{
printf("\n你的输入有误，请重新输入");
scanf("%d",&changenum);
}
if(changenum!=0)
printf("\n下面请输入更新边的信息：\n");
for(i=1;i<=changenum;i++)
{
printf("\n修改的第%d条边的起点 终点 长度为:",i);
scanf("%d %d %d",&v0,&v1,&distance);
l=locatevet(campus,v0);
n=locatevet(campus,v1);
if(l>=0&&n>=0)
{
m->mat[l][n].wet=distance;
m->mat[n][l].wet=m->mat[l][n].wet;
}
}
return 1;
}
//9.增加一条边
int incside(mapstr *m)
{
int l,n,distance;
printf("\n请输入边的起点和终点编号，权值:");
scanf("%d%d%d",&l,&n,&distance);
while(l<0||l>m->vetnum||n<0||n>m->vetnum)
{
printf("输入错误，请重新输入");
scanf("%d %d",&l,&n);
}
if(locatevet(campus,l)<0)
{
printf("此节点%d已删除",l);
return 1;
}
if(locatevet(campus,n)<0)
{
printf("此节点%d已被删除",n);
return 1;
}
m->mat[l][n].wet=distance;
m->mat[n][l].wet=m->mat[l][n].wet;
m->sidenum ++;
return 1;
}
//10.增加一个结点
int incvet(mapstr *m)
{
int i;
m->vetnum++;//顶点数加一
printf("请输入您要增加结点的信息：");
printf("\n编号:");
scanf("%d",&m->vets[m->vetnum].number);
printf("名称:");
scanf("%s",&m->vets[m->vetnum].name);
printf("简介:");
scanf("%s",&m->vets[m->vetnum].intro);
for(i=1;i<=m->vetnum;i++)
{
m->mat[m->vetnum][i].wet=Infinity;
m->mat[i][m->vetnum].wet=Infinity;
}
return 1;
}
//11.删除一个结点
int delvet(mapstr *m)
{
int i=0,j,l,v;
if(m->vetnum<=0)
{
printf("图中已无顶点");
return 1;
}
printf("\n下面请输入您要删除的景点编号：");
scanf("%d",&v);
while(v<0||v>m->vetnum)
{
printf("\n输入错误！请重新输入:");
scanf("%d",&v);
}
l=locatevet(campus,v);
if(l<0)
{
printf("顶点%d已删除\n",v);
return 1;
}
for(i=l;i<=m->vetnum-1;i++)
for(j=1;j<=m->vetnum;j++)//将二维数组中的第m+1行依次向前移动一行（删除第m行）
m->mat[i][j]=m->mat[i+1][j];
for(i=l;i<=m->vetnum-1;i++)
for(j=1;j<=m->vetnum;j++)//将二维数组中的第m+1列依次向前移动一列（删除第m列）
m->mat[j][i]=m->mat[j][i+1];
m->vets[v].number=-1;//表示此点已删除，后期打印也不会显示该点
m->vetnum--;//顶点个数-1
return 1;
}
//12.删除一条边
int delside(mapstr *m)
{
int l,n,v0,v1;
if(m->vetnum<=0)
{
printf("图中没有边了，无法删除");
return 1;
}
printf("\n下面请输入您要删除的边的起点和终点编号：");
scanf("%d %d",&v0,&v1);
l=locatevet(campus,v0);
if(m<0)
{
printf("此%d顶点已删除",v0);
return 1;
}
n=locatevet(campus,v1);
if(n<0)
{
printf("此%d顶点已删除",v1);
return 1;
}
m->mat[l][n].wet=Infinity;//将删掉的边的权值改为无穷
m->mat[n][l].wet=Infinity;
m->sidenum--;//图的边数减一
return 1;
}
//13.输出图的邻接矩阵的值
void printmapstr(mapstr m)
{
int i,j,k=0;
for(i=1;i<=vertex;i++)
{
if(m.vets[i].number!=-1)
printf("%6d",i);
}
printf("\n");
for(i=1;i<=m.vetnum;i++)
{
for(j=1;j<=m.vetnum;j++)
{
if(m.mat[i][j].wet==Infinity)
printf(" **** ");
else
printf("%6d",m.mat[i][j].wet);
k++;
if(k%m.vetnum==0)
printf("\n");
}
}
}
//14.图的操作主函数
int changemap(mapstr *m)
{
int choice;
printf("(1)重新建图 (2)删除结点 (3)删除边\n");
printf("(4)增加结点 (5)增加边 (6)更新信息\n");
printf("(7)输出邻接矩阵 (8)返回主页 \n");
do
{
printf("请输入你的选择：");
scanf("%d",&choice);
switch(choice)
{
case 1:creatmap(m);break;
case 2:delvet(m);break;
case 3:delside(m);break;
case 4:incvet(m);break;
case 5:incside(m);break;
case 6:newmap(m);break;
case 7:printmapstr(campus);break;
case 8:system("cls");menu();return 1;
default:printf("未找到该功能，请输入有效选项！\n");break;
}
}while(choice);
}
//15.用户登录
int userlog()
{
int i;
int a[6]={
1,2,3,4,5,6},b[6];
printf("\n请输入六位密码(用空格隔开)：\n");
for(i=0;i<6;i++)
scanf("%d",&b[i]);
for(i=0;i<6;i++)
{
if(a[i]!=b[i])
{
printf("密码错误！自动返回主页面\n");
menu();
return 0;
}
}
printf("密码正确，登陆成功！\n\n");
changemap(&campus);
}
//16.弗洛伊德算法
void floyd(mapstr m)
{
int i,j,k;
for(i=1;i<=vertex;i++)//将图的邻接矩阵赋值给 shortest二维数组，将矩阵pathh全部初始化为-1
{
for(j=1;j<=vertex;j++)
{
shortest[i][j]=m.mat[i][j].wet;
pathh[i][j]=j;
}
}
int ii,jj,k1=0;
for(ii=1;ii<=vertex;ii++)
printf("%6d",ii);//横着标号1到11
printf("\n");
for(ii=1;ii<=vertex;ii++)
{
printf("%d",ii);
for(jj=1;jj<=vertex;jj++)
{
printf("%6d",pathh[ii][jj]);
k1++;
if(k1%vertex==0)
printf("\n");
}
}
printf("\n\n\n");
for(k=1;k<=vertex;k++)//核心操作，完成了以k为中间点对所有的顶点对（i,j）进行检测和修改
{
for(i=1;i<=vertex;i++)
{
for(j=1;j<=vertex;j++)
{
if(shortest[i][j]>shortest[i][k]+shortest[k][j])
{
shortest[i][j]=shortest[i][k]+shortest[k][j];
pathh[i][j]=pathh[i][k];//记录一下所走的路 //P数组用来存放前驱顶点
}
}
}
}
}
//17.输出数组
void printarray()
{
int i,j,k=0;
for(i=1;i<=vertex;i++)
printf("%6d",i);//横着标号0-11
printf("\n");
for(i=0;i<=vertex;i++)
{
printf("%d",i);//竖着标号0-11
for(j=1;j<=vertex;j++)
{
printf("%6d",pathh[i][j]);
k++;
if(k%vertex==0)
printf("\n");
}
}
printf("\n\n\n");
}
//18.输出最短路径
void display(mapstr m,int i,int j)
{
int a,b;
a=i,b=j;
printf("您要查询的两景点间最短路径：\n\n");
printf("%d%s",a,m.vets[a].name);
while(pathh[i][j]!=b)
{
printf("-->%d%s",pathh[i][j],m.vets[pathh[i][j]].name);
i=pathh[i][j];
}
printf("-->%d%s\n\n",b,m.vets[b].name);
printf("%s-->%s的最短路径是：%d米\n\n",m.vets[a].name,m.vets[b].name,shortest[a][b]);
}
//19任意两点间距离（16-19）
int shortdistance(mapstr m)
{
int i,j;
printf("请输入要查询的两个景点的数字编号(用空格隔开)\n");
scanf("%d %d",&i,&j);
if(i<0||i>vertex||j<0||j>vertex)
{
printf("输入信息有误！\n\n");
printf("请输入要查询的两个景点的数字编号(用空格隔开)\n");
scanf("%d %d",&i,&j);
}
else
{
floyd(m);
printarray();
display(m,i,j);
}
return 1;
}
//20显示所有景点信息
void compusinfor(mapstr m)
{
int i;
printf(" \n\n编号 景点名称 简介\n");
printf("*************************************************************************\n");
for(i=1;i<=vertex;i++)
{
if(m.vets[i].number!=-1)
printf("%-10d%-25s%-70s\n",m.vets[i].number,m.vets[i].name,m.vets[i].intro);
}
printf("*************************************************************************\n");
}
//21青岛农业大学地图
void schoolmap()
{
printf(" ---------------------------------青岛农业大学地图----------------------------------------\n");
printf("````````````````````````````````````````````东苑``````````````````````````````````````````\n");
printf("````````````````````````````````````````````宿舍``````````````````````````````````````````\n");
printf("````````````````````````````````````````````````````````````````````````````东苑``````````\n");
printf("````````````````````````````````````````````````````````````````````````````餐厅``````````\n");
printf("```````````````````````````````体育场`````````````````````````````````````````````````````\n");
printf("``````````````````````````````````````````````````````````````````````````````````````````\n");
printf("````````````````````````````````````````````农大花园``````````````````````足球场``````````\n");
printf("``````````````````````文体馆``````````````````````````````````````````````````````````````\n");
printf("``````````````````````````````````````````````````````````````````````````````````````````\n");
printf("``````````````````````````````````````````````````````````````````````````````````````````\n");
printf("西苑`西苑``地下````````````````````````````````````````虹子湖`````````````````````````````\n");
printf("宿舍`餐厅``通道```````````````````````````````````````````````````````````````````````````\n");
printf("```````````````````````````````````````````````````````````````````````````工程楼````````\n");
printf("``````````````科技楼``````````````````````````````````````````````````````````````````````\n");
printf("``````````````信息楼```````````生物楼`````办公楼````教学楼`````````````````文经楼```````\n");
printf("``````````````化学楼``````````````````````图书馆``````````````````````````````````````````\n");
printf("``````````````````````````````````````````````````````````````````````````````````````````\n");
printf("`````````````````````````````````````````虹子广场```````````````````````````````````````````\n");
printf("\n\n");
printf("\n 西<---|--->东 \n");
}
//22用户界面
void mainwork()
{
menu();
int choice;
campus=initmap();
do
{
printf("请输入你的选择：");
scanf("%d",&choice);
switch(choice)
{
case 1:system("cls");menu();break;
case 2:system("cls");shortestpath(campus);break;
case 3:system("cls");shortdistance(campus);break;
case 4:system("cls");allpath(campus);break;
case 5:system("cls");compusinfor(campus);break;
case 6:system("cls");schoolmap();break;
case 7:system("cls");userlog();break;
case 0:system("cls");printf("谢谢使用\n");break;
default:printf("未找到该功能，请输入有效选项！\n");break;
}
}while(choice);
}
int main()
{
mainwork();
return 0;
}
5 program results
Figure 5-1 Program home page
6 Summary
In the course design of data structure, the campus tour guide system of Qingdao Agricultural University has been completed. For the realization of campus modernization, visitors and freshmen who visit our school can more easily understand the school's scenic spots, facilitate visits, and reduce the number of tour guides. I wrote this campus guide system to provide a campus guide system for freshmen or those who visit the school for the first time to guide people to visit. The campus tour guide system is solved by using the graph structure in the data structure. It uses the campus attractions as the nodes of the graph, the paths between the scenic spots as the edges of the graph, and the path distance as the weight of the edges.
Disadvantages:
1: The campus map written with the printf statement is too blunt, not beautiful and real.
2: There is a gap with the real navigation system, and the real-time path problem cannot be reflected and solved.
7 Harvesting and experience
Through this data structure course design, I have fully learned the relevant knowledge of the graph in the data structure, especially the method of finding the shortest path. The first is the single-source shortest path (using the Dijkstra algorithm), and the second is any Vertex shortest path (Floyd algorithm), master the core ideas of these two algorithms, watch the teaching videos repeatedly during the learning process, and then read the books to carefully analyze the characteristics of these algorithms to understand the essential content of these two algorithms.
After choosing the topic of the campus tour guide system, I did not rush to do it, but thought about it first, from the storage structure to the structure, to the algorithm used, and the function of the project. When encountering problems, face them bravely, not be afraid of problems, overcome difficulties, and let yourself really learn knowledge. In completing the course design of the data structure, I felt the joy of applying the acquired knowledge to practice, consolidated the knowledge points that were not firmly memorized in the classroom, and realized the joy of learning!
Through the course design, I have deepened my understanding and consolidation of the theoretical knowledge I have learned, and can apply the pure theory in the textbook to practice, further deepening my understanding of knowledge. At the same time, it also helps to understand other knowledge, which gives me a deeper understanding of data structures.
8 References
[1] Data Structure Tutorial Tsinghua University Press Li Chunbao edited 2017.5
[2] Bilibili Data Structure and Algorithm Video Small Turtle
[3] Bilibili Data Structure and Algorithm Basic Video Qingdao University Wang Zhuo
[4] Baidu Netdisk Data structure video kingly series
The course design is hard-won, if you think it is useful, give me a thumbs up to encourage me, thank you~