南邮数据结构实验三.doc

.实 验 报 告（2015 / 2016 学年 第 一 学期）课程名称数据结构实验名称图的基本运算及飞机换乘次数最少问题实验时间2015年12月4日指导单位计算机科学与技术系指导教师黄海平学生姓名陈明阳班级学号Q14010119学院(系)贝尔英才专 业信息科技强化班;.实 验 报 告实验名称图的基本运算及飞机换乘次数最少问题指导教师黄海平实验类型验证实验学时4实验时间12.4一、 实验目的和要求飞机最少换乘次数问题。（1）设有n个城市，编号为0n-1，m条航线的起点和终点由用户输入提供。寻找一条换乘次数最少的线路方案。（2）参考：可以使用有向图表示城市间的航线；只要两城市间有航班，则图中这两点间存在一条权值为1的边；可以使用Dijkstra算法实现。二、实验环境(实验设备)VSUAL STUDIO2015三、实验原理及内容对象视图：源代码：Graph.h#include #includeusing namespace std;const int INF = 2147483647;enum ResultCode Underflow, Duplicate, Failure, Success, NotPresent, OutOfBounds ;template class Graph /抽象类 public:virtual ResultCode Insert(int u, int v, T w) = 0;virtual ResultCode Remove(int u, int v) = 0;virtual bool Exist(int u, int v)const = 0;protected:int n, e;template class MGraph :public Graph /邻接矩阵类 public: MGraph(int mSize, const T noedg); MGraph(); ResultCode Insert(int u, int v, T w); ResultCode Remove(int u, int v); bool Exist(int u, int v)const; int Choose(int *d, bool *s); void Dijkstra(int v, T *d, int *path);protected: T *a; T noEdge;template MGraph:MGraph(int mSize, const T noedg) n = mSize; e = 0; noEdge = noedg; a = new T*n; for (int i = 0; in; i+) ai = new Tn; for (int j = 0; jn; j+) aij = noEdge; aii = 0; template MGraph:MGraph() for (int i = 0; in; i+) deleteai; deletea;template ResultCode MGraph:Insert(int u, int v, T w) if (u0 | vn - 1 | vn - 1 | u = v) return Failure; if (auv != noEdge) return Duplicate; auv = w; avu = w; e+; return Success;template ResultCode MGraph:Remove(int u, int v) if (u0 | vn - 1 | vn - 1 | u = v) return Failure; if (auv = noEdge) return NotPresent; auv = noEdge;avu = noEdge; e-; return Success;templatebool MGraph:Exist(int u, int v)const if (u0 | vn - 1 | vn - 1 | u = v | auv = noEdge) return false; return true;template int MGraph:Choose(int *d, bool *s) /求最小di int i, minpos; T min; min = INF; minpos = -1; for (i = 0; in; i+) if (di = min&!si) min = di; minpos = i; return minpos;template void MGraph:Dijkstra(int v, T *d, int *path) /迪杰斯特拉算法 int i, k, w; if (vn - 1) throw OutOfBounds; bool *s = new booln; for (i = 0; in; i+) si = false; di = avi; if (i != v&diINF) pathi = v; else pathi = -1; sv = true; dv = 0; for (i = 1; in; i+) k = Choose(d, s); sk = true; for (w = 0; wn; w+) if (!sw & (dk + akw)dw) dw = dk + akw; pathw = k; 源.cpp#include #include#includeGraph.husing namespace std;int main() int n, m; cout n; cout m; MGraphA(n, INF); int c, f; cout 请输入每条航线的起点和终点: endl; for (int i = 0; im; i+) cout 航线 i + 1 c f; A.Insert(c, f, 1); char s; do int v, w; cout v w; while (v0 | wn - 1 | vn - 1) cout v w; int *b = new intn; int *d = new intn; int *path = new intn; A.Dijkstra(v, d, path); int e = n - 1;int j, i,k = 0; for (j = 0; jn; j+) bj = -2; if (w != v) j = w; while (pathj != -1) be = pathj; e-; j = pathj; if (e = n - 1 | dj = INF) cout 这两点间无线路! endl; else cout 从 v 城市到 w 城市的换乘次数最小的线路方案为:; for (k = 0; kn; k+) if (bk != -2) cout bk ,; cout w endl; if (w = v) cout 看来你钱比较多 endl; deleteb; deleted; deletepath; cout s; while (s != Y&s != y&s != n&s != N) cout s; while (s = Y | s = y); return 0;运行截图：实 验 报 告四、实验小结（包括问题和解决方法、心得体会、意见与建议等）使用邻接表表示的图和迪杰斯特拉算法实现飞机最小换乘路径本来不算一件难事，这个程序实现起来的重点在于容错性的问题。在考虑到输入时可能有的各种各样问题时主程序中的输入判断就要多改进。还有就是我怎么都觉得飞机路线图怎么