问题描述：
一农夫带着一头狼，一只羊和一个白菜过河，小船只能一次装载农夫和一样货物，狼会吃羊，羊会吃白菜，只有农夫在时才安全。现欲让所有物品包括农夫都安全过道河对岸，求最佳算法。
解法如下：
1.农夫带羊过去，自己回来
2.农夫带狼过去，带羊回来
3.农夫带白菜过去，自己回来
4.农夫带羊过去
全部安全过岸.
深度优先搜索方法：首先扩展最新产生的结点，每层只对一个结点进行扩展，除非搜索失败或以达到预先约定的最大深度，才会退回去搜索原来来忽略的结点。
广度优先搜索方法：以接近起始结点的程度依次扩展结点，即对下一层结点搜索前，必须先搜索完本层所有结点。
深度优先（栈）源代码：
#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
char* 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;
}
输出结果：
广度优先（队列）源代码：
#include
#include
#define MAXNUM 20
typedef int DataType;
struct SeqQueue {
int f, r;
DataType q[MAXNUM];
};
typedef struct SeqQueue *PSeqQueue;
PSeqQueue createEmptyQueue_seq( void
) {
PSeqQueue paqu = (PSeqQueue)malloc(sizeof(struct SeqQueue));
if (paqu == NULL)
printf("Out of space!! \n");
else
paqu->f = paqu->r = 0;
return (paqu);
}
int isEmptyQueue_seq( PSeqQueue paqu ) {
return paqu->f == paqu->r;
}
void enQueue_seq( PSeqQueue paqu, DataType x ) {
if ( (paqu->r + 1) % MAXNUM == paqu->f )
printf( "Full queue.\n" );
else {
paqu->q[paqu->r] = x;
paqu->r = (paqu->r + 1) % MAXNUM;
}
}
void deQueue_seq( PSeqQueue paqu ) {
if( paqu->f == paqu->r )
printf( "Empty Queue.\n" );
else
paqu->f = (paqu->f + 1) % MAXNUM;
}
DataType frontQueue_seq( PSeqQueue paqu ) {
return (paqu->q[paqu->f]);
}
int farmer(int location) {
return 0 != (location & 0x08);
}
int wolf(int location) {
return 0 != (location & 0x04);
}
int cabbage(int location) {
return 0 != (location & 0x02);
}
int goat(int location) {
return 0 !=(location & 0x01);
}
int safe(int location) {
if ((goat(location) == cabbage(location)) &&
(goat(location) != farmer(location)) )
return 0;
if ((goat(location) == wolf(location)) &&
(goat(location) != farmer(location)))
return 0;
return 1;
}
void farmerProblem( ) {
int movers, i, location, newlocation;
int route[16];
PSeqQueue moveTo;
moveTo = createEmptyQueue_seq( );
enQueue_seq(moveTo, 0x00);
for (i = 0; i < 16; i++) route[i] = -1;
route[0]=0;
while (!isEmptyQueue_seq(moveTo)&&(route[15] == -1)) {
location = frontQueue_seq(moveTo);
deQueue_seq(moveTo);
for (movers = 1; movers <= 8; movers <<= 1) {
if ((0 != (location & 0x08)) == (0 != (location & movers))) {
newlocation = location^(0x08|movers);
if (safe(newlocation) && (route[newlocation] == -1)) {
route[newlocation] = location;
enQueue_seq(moveTo, newlocation);
}
}
}
}
if(route[15] != -1) {
printf("The reverse path is : \n");
for(location = 15; location >= 0; location = route[location]) {
printf("The location is : %d\n",location);
if (location == 0) return;
}
}
else
printf("No solution.\n");
}
int main() {
farmerProblem( );
return 0;
}