数据结构问题 求大神带

Question

求数据结构的迷宫问题和农夫过河问题 和一元稀疏矩阵的加减法问题 急

Answer 1

正好我今天也是那个课设
农夫过河
#include
#include
using namespace std;
#define VertexNum 16 //最大顶点数
typedef struct // 图的顶点
{
int farmer; // 农夫
int wolf; // 狼
int sheep; // 羊
int veget; // 白菜
}Vertex;
typedef struct
{
int vertexNum; // 图的当前顶点数
Vertex vertex[VertexNum]; // 顶点向量（代表顶点）
bool Edge[VertexNum][VertexNum]; // 邻接矩阵. 用于存储图中的边，其矩阵元素个数取决于顶点个数，与边数无关
}AdjGraph; // 定义图的邻接矩阵存储结构
bool visited[VertexNum] = {false}; // 对已访问的顶点进行标记（图的遍历）
int retPath[VertexNum] = {-1}; // 保存DFS搜索到的路径，即与某顶点到下一顶点的路径
// 查找顶点（F，W，S，V）在顶点向量中的位置
int locate(AdjGraph *graph, int farmer, int wolf , int sheep, int veget)
{
// 从0开始查找
for (int i = 0; i < graph->vertexNum; i++)
{
if ( graph->vertex[i].farmer == farmer && graph->vertex[i].wolf == wolf
&& graph->vertex[i].sheep == sheep && graph->vertex[i].veget == veget )
{
return i; //返回当前位置
}
}
return -1; //没有找到此顶点
}
// 判断目前的（F，W，S，V）是否安全
bool isSafe(int farmer, int wolf, int sheep, int veget)
{
//当农夫与羊不在一起时，狼与羊或羊与白菜在一起是不安全的
if ( farmer != sheep && (wolf == sheep || sheep == veget) )
{
return false;
}
else
{
return true; // 安全返回true
}
}
// 判断状态i与状态j之间是否可转换
bool isConnect(AdjGraph *graph, int i, int j)
{
int k = 0;
if (graph->vertex[i].wolf != graph->vertex[j].wolf)
{
k++;
}
if (graph->vertex[i].sheep != graph->vertex[j].sheep)
{
k++;
}
if (graph->vertex[i].veget != graph->vertex[j].veget)
{
k++;
}
// 以上三个条件不同时满足两个且农夫状态改变时，返回真, 也即农夫每次只能带一件东西过桥
if (graph->vertex[i].farmer != graph->vertex[j].farmer && k <= 1)
{
return true;
}
else
{
return false;
}
}
// 创建连接图
void CreateG(AdjGraph *graph)
{
int i = 0;
int j = 0;
// 生成所有安全的图的顶点
for (int farmer = 0; farmer <= 1; farmer++)
{
for (int wolf = 0; wolf <= 1; wolf++)
{
for (int sheep = 0; sheep <= 1; sheep++)
{
for (int veget = 0; veget <= 1; veget++)
{
if (isSafe(farmer, wolf, sheep, veget))
{
graph->vertex[i].farmer = farmer;
graph->vertex[i].wolf = wolf;
graph->vertex[i].sheep = sheep;
graph->vertex[i].veget = veget;
i++;
}
}
}
}
}
// 邻接矩阵初始化即建立邻接矩阵
graph->vertexNum = i;
for (i = 0; i < graph->vertexNum; i++)
{
for (j = 0; j < graph->vertexNum; j++)
{
// 状态i与状态j之间可转化，初始化为1，否则为0
if (isConnect(graph, i, j))
{
graph->Edge[i][j] = graph->Edge[j][i] = true;
}
else
{
graph->Edge[i][j] = graph->Edge[j][i] = false;
}
}
}
return;
}
// 判断在河的那一边
string judgement(int state)
{
return (0 == state) ? "左岸" : "右岸" ;
}
// 输出从u到v的简单路径，即顶点序列中不重复出现的路径
void printPath(AdjGraph *graph, int start, int end)
{
int i = start;
cout << "farmer" << ", wolf" << ", sheep" << ", veget" << endl;
while (i != end)
{
cout << "(" << judgement(graph->vertex[i].farmer) << ", " << judgement(graph->vertex[i].wolf)
<< ", " << judgement(graph->vertex[i].sheep) << ", " << judgement(graph->vertex[i].veget) << ")";
cout << endl;
i = retPath[i];
}
cout << "(" << judgement(graph->vertex[i].farmer) << ", " << judgement(graph->vertex[i].wolf)
<< ", " << judgement(graph->vertex[i].sheep) << ", " << judgement(graph->vertex[i].veget) << ")";
cout << endl;
}
// 深度优先搜索从u到v的简单路径 //DFS--Depth First Search
void dfsPath(AdjGraph *graph, int start, int end)
{
int i = 0;
visited[start] = true; //标记已访问过的顶点
if (start == end)
{
return ;
}
for (i = 0; i < graph->vertexNum; i++)
{
if (graph->Edge[start][i] && !visited[i])
{
retPath[start] = i;
dfsPath(graph, i, end);
}
}
}
int main()
{
AdjGraph graph;
CreateG(&graph);
int start = locate(&graph, 0, 0, 0, 0);
int end = locate(&graph, 1, 1, 1, 1);
dfsPath(&graph, start, end);
if (visited[end]) // 有结果
{
printPath(&graph, start, end);
return 0;
}
return -1;
}