我找到了，关于农夫过河问题，仍然谢谢你们
“数据结构”部分
农夫过河问题描述的是一个农夫带着一只狼、一只羊和一棵白菜，处于河的南岸，而他要将这些东西带到北岸。但是他面前只有一条小船，而每次他只能带一件东西过河，由于食物链的关系，他不能将狼和羊同时留下，也不能将羊和白菜同时留下，但可以将狼和白菜同时留下。编写一段代码，描述农夫所采用的过河方法。
/* 用队列解决农夫过河问题的算法*/
#include<stdio.h>
#include<stdlib.h>
#define MAXNUM 20
typedef int DataType;
struct SeqQueue { /* 顺序队列类型定义 */
int f, r;
DataType q[MAXNUM];
};
typedef struct SeqQueue *PSeqQueue; /* 顺序队列类型的指针类型 */
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;
}
/* 在队列中插入一元素x */
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;
}
}
/* 删除队列头部元素 */
void deQueue_seq( PSeqQueue paqu ) {if( paqu->f == paqu->r )
printf( "Empty Queue.\n" );
else
paqu->f = (paqu->f + 1) % MAXNUM;
}
/* 对非空队列,求队列头部元素 */
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);
}
/* 若状态安全则返回true */
int safe(int location) {/* 羊吃白菜 */
if ((goat(location) == cabbage(location)) &&
(goat(location) != farmer(location)) )
return 0;
/* 狼吃羊 */
if ((goat(location) == wolf(location)) &&
(goat(location) != farmer(location)))
return 0;
return 1; /* 其他状态是安全的 */
}
void farmerProblem( ) {int movers, i, location, newlocation;
int route[16]; /*记录已考虑的状态路径*/
PSeqQueue moveTo;
/*准备初值*/
moveTo = createEmptyQueue_seq( );
enQueue_seq(moveTo, 0x00);
for (i = 0; i < 16; i++) route[i] = -1;
route[0]=0;
/*开始移动*/
while (!isEmptyQueue_seq(moveTo)&&(route[15] == -1)) {/*得到现在的状态*/
location = frontQueue_seq(moveTo);
deQueue_seq(moveTo);
for (movers = 1; movers <= 8; movers <<= 1) {/* 农夫总是在移动，随农夫移动的也只能是在农夫同侧的东西 */
if ((0 != (location & 0x08)) == (0 != (location & movers))) {newlocation = location^(0x08|movers);
if (safe(newlocation) && (route[newlocation] == -1)) { route[newlocation] = location;
enQueue_seq(moveTo, newlocation);
}
}
}
}
/* 打印出路径 */
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;
}
[此贴子已经被作者于2005-12-22 15:07:26编辑过]
