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@ -123,53 +123,125 @@ typedef struct
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## 3.1 核心数据结构的实现
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利用循环结构实现查找某一个城市的信息
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如下:
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int search(MatGrath &G)
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{ int a;
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int flag=1;
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printf("请输入您要查询的城市编号\n");
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scanf("%d",&a);
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while(flag)
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{
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if(a<=0||a>G.vexnum)
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{
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printf("编号不存在,请重新输入\n");
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scanf("%d",&a);
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}
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else
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{ flag=0;
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核心算法Dijkstra
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算法floyd
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项目文件结构可以设计为:
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```
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CityRoadNav/ - CityRoadNav.vcxproj // Visual Studio项目文件 - CityRoadNav.vcxproj.filters - CityRoadNav.slnSource Files/ // 源代码目录 - CityNetwork.cpp // 城市网络类,存储城市数据及邻接矩阵 - CityNetwork.h - Dijkstra.cpp // Dijkstra算法类,实现最短路径搜索 - Dijkstra.h - DynamicProgramming.cpp // 动态规划算法类,实现最优路径搜索 - DynamicProgramming.h - main.cpp // 主程序 - ...Header Files/ // 头文件目录 - ... Resources/ // 资源文件目录 - CityInfo.txt // 城市信息及邻接矩阵
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```
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如以下代码,采用迪杰斯特拉算法实现求取路径最短值
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```
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int Ppath2(MatGrath &G,int path[],int i,int v) //前向递归查找路径上的顶点
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{
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int k;
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k=path[i];
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if (k==v)
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return k; //找到了起点则返回
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Ppath2(G,path,k,v); //找顶点k的前一个顶点
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printf("%s->",G.vexs[k].sight);//输出顶点k
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}
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printf("编号 景点名称 简介 \n");
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printf(" %-4d %-16s%-58s\n",G.vexs[a].no,G.vexs[a].sight,G.vexs[a].introduction);
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;
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}
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int danyuan(MatGrath &G,int v)//求两点之间的最短路径
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{
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int dist[MAXV],path[MAXV];
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int s[MAXV];
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int mindis,i,j,u;
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for (i=1; i<=G.vexnum; i++)
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{
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dist[i]=G.arc[v][i].length; //距离初始化
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s[i]=0; //s[]置空
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if (G.arc[v][i].length<INF) //路径初始化
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path[i]=v;
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else
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path[i]=-1;
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}
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s[v]=1;
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path[v]=0; //源点编号v放入s中
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for (i=1; i<=G.vexnum; i++) //循环直到所有顶点的最短路径都求出
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{
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mindis=INF; //mindis置最小长度初值
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for (j=1; j<=G.vexnum; j++) //选取不在s中且具有最小距离的顶点u
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if (s[j]==0 && dist[j]<mindis)
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{
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u=j;
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mindis=dist[j];
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}
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return 0;
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s[u]=1; //顶点u加入s中
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for (j=1; j<=G.vexnum; j++) //修改不在s中的顶点的距离
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if (s[j]==0)
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if (G.arc[u][j].length<INF && dist[u]+G.arc[u][j].length<dist[j])
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{
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dist[j]=dist[u]+G.arc[u][j].length;
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path[j]=u;
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}
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}
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for(i=1; i<=G.vexnum; i++)
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if (s[i]==1&&v!=i)
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{
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printf(" 从%s到%s的最短路径长度为:%d米\t路径为:",G.vexs[v].sight,G.vexs[i].sight,dist[i]);
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printf("%s->",G.vexs[v].sight); //输出路径上的起点
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Ppath2(G,path,i,v); //输出路径上的中间点
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printf("%s\n",G.vexs[i].sight); //输出路径上的终点
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}
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利用深度遍历实现输出城市信息
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如下:
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bool visited[MAXV];
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int DFS(MatGrath &G,int m)//深度优先搜索
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}
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int Ppath1(MatGrath &G,int path[][MAXV],int v,int w) //前向递归查找路径上的顶点
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{
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int k;
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k=path[v][w];
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if (k==-1) return 0; //找到了起点则返回
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Ppath1(G,path,v,k); //找顶点i的前一个顶点k
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printf("%s->",G.vexs[k].sight);
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Ppath1(G,path,k,w); //找顶点k的前一个顶点j
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}
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```
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如下采用floyd算法求给出的两点之间的最少费用
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```
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void ShortestMoney(MatGrath &G,int v,int w)//求花费最少的路径
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{
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int A[MAXV][MAXV],path[MAXV][MAXV];
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int i,j,k;
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for (i=0; i<G.vexnum; i++)
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for (j=0; j<G.vexnum; j++)
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{
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int i;
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printf("%d %s %s\n",G.vexs[m].no,G.vexs[m].sight,G.vexs[m].introduction);
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visited[m]=true;
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for(i=1;i<=G.vexnum;i++)
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{
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if(G.arc[m][i].length!=INF&&visited[i]==false)
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A[i][j]=G.arc[i][j].money;
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path[i][j]=-1; //i到j没有边
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}
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for (k=0; k<G.vexnum; k++)
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{
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for (i=0; i<G.vexnum; i++)
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for (j=0; j<G.vexnum; j++)
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if (A[i][j]>A[i][k]+A[k][j])
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{
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DFS(G,i);
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A[i][j]=A[i][k]+A[k][j];
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path[i][j]=k;
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}
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}
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}
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return 0;
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}
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if (A[v][w]==INF)
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{
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if (v!=w)
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printf("从%d到%d没有路径\n",v,w);
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}
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else
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{
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printf(" 从%s到%s路径费用:%d元人民币 路径:",G.vexs[v].sight,G.vexs[w].sight,A[v][w]);
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printf("%s->",G.vexs[v].sight); //输出路径上的起点
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Ppath1(G,path,v,w); //输出路径上的中间点
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printf("%s\n",G.vexs[w].sight); //输出路径上的终点
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}
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}
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```
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## 3.2 核心算法的实现
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