农夫过河

}
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" );
// 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++)
{
graph->vertex[i].farmer = farmer;
graph->vertex[i].wolf = wolf;
graph->vertex[i].sheep = sheep;
}
void enQueue_seq( PSeqQueue paqu, DataType x ) {
if ( (paqu->r + 1) % MAXNUM == paqu->f )
printf( "Full queue.\n" );
else {
paqu->q[paqu->r] = x;
{
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，Fra bibliotek）是否安全 bool isSafe(int farmer, int wolf, int sheep, int veget)
{
while (i != end)
{
cout << "(" << judgement(graph->vertex[i].farmer) << ", " << judgement(graph->vertex[i].wolf)
<< ", " << judgement(graph->vertex[i].sheep) << ", " << judgement(graph->vertex[i].veget) << ")";
{
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))
#include<stdio.h>
#include<stdlib.h>
#define MAXNUM 20
typedef int DataType;
struct SeqQueue {
int f, r;
DataType q[MAXNUM];
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 ( graph->vertex[i].farmer == farmer && graph->vertex[i].wolf == wolf
&& graph->vertex[i].sheep == sheep && graph->vertex[i].veget == 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) << ")";
}
// 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;
if (start == end)
{
return ;
}
for (i = 0; i < graph->vertexNum; i++)
{
if (graph->Edge[start][i] && !visited[i])
{
retPath[start] = i;
{
k++;
}
if (graph->vertex[i].sheep != graph->vertex[j].sheep)
{
k++;
}
if (graph->vertex[i].veget != graph->vertex[j].veget)
{
k++;
}
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：
paqu->r = (paqu->r + 1) % MAXNUM;
}
}
void deQueue_seq( PSeqQueue paqu ) {
if( paqu->f == paqu->r )
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<iostream>
printf("Out of space!! \n");
else
paqu->f = paqu->r = 0;
return (paqu);
}
int isEmptyQueue_seq( PSeqQueue paqu ) {
return paqu->f == paqu->r;
// 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
//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
}
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);
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。