#include <stdio.h>
#include <stdlib.h>
#define maxloop 100
//最大层数，对于不同的扩展方法自动调整取值
#define pristnum 3
#define slavenum 3
struct SPQ
{
int sr,pr;
//船运行一个来回后河右岸的野人、传教士的人数
int sl,pl;
//船运行一个来回后河左岸的野人、传教士的人数
int ssr,spr;
//回来（由左向右时）船上的人数
int sst,spt;
//去时（由右向左时）船上的人数
int loop;
//本结点所在的层数
struct SPQ *upnode ,*nextnode;//本结点的父结点和同层的下一个结点的地址
}spq;
int loopnum;//记录总的扩展次数
int openednum;//记录已扩展节点个数
int unopenednum;//记录待扩展节点个数
int resultnum;
struct SPQ *opened;
struct SPQ *oend;
struct SPQ *unopened;
struct SPQ *uend;
struct SPQ *result;
void initiate();
void releasemem();
void showresult();
void addtoopened(struct SPQ *ntx);
int search();
void goon();
int stretch(struct SPQ* ntx);
void recorder();
void main()
{
int flag;
//标记扩展是否成功
for( ; ; )
{
initiate();
flag = search ();
if(flag == 1)
{
recorder();
releasemem();
showresult();
goon();
}
else
{
printf("无法找到符合条件的解");
releasemem();
goon();
}
}
}
void initiate()
{
int x;
char choice;
uend = unopened = (struct SPQ*)malloc(sizeof(spq));
if(uend==NULL)
{
printf("\n内存不够！\n");
exit(0);
}
unopenednum=1;
openednum=0;
unopened -> upnode = unopened;
//保存父结点的地址以成链表
unopened -> nextnode = unopened;
unopened -> sr = slavenum;
unopened -> pr = pristnum;
unopened -> sl = 0;
unopened -> pl = 0;
unopened -> sst = 0;
unopened -> spt = 0;
unopened -> ssr = 0;
unopened -> spr = 0;
unopened -> loop = 0;
printf("题目：设有n个传教士和m个野人来到河边，打算乘一只船从右岸到左岸去。\n");
printf("该船的负载能力为两人。在任何时候，如果野人人数超过传教士人数，野人\n");
printf("就会把传教士吃掉。他们怎样才能用这条船安全的把所有人都渡过河去?\n");
printf("\n默认的n、m值皆为3\n");
for(;;)
{
printf("\n是否修改？(Y/N)");
scanf("%s",&choice);
choice=toupper(choice);
if(choice=='Y')
{
printf("\n请输入传教士人数");
for(;;)
{
scanf("%d",&x);
if(x>0)
{
unopened -> pr = x;
break;
}
else printf("\n输入值应大于0！\n请重新输入");
}
printf("\n请输入野人人数");
for(;;)
{
scanf("%d",&x);
if(x>0)
{
unopened -> sr = x;
break;
}
else printf("\n输入值应大于0！\n请重新输入");
}
break;
}
if(choice=='N')break;
}
}
int search()
{
int flag;
struct SPQ *ntx;
//提供将要扩展的结点的指针
for( ; ; )
{
ntx = unopened;
//从待扩展链表中提取最前面的一个
if(ntx->loop == maxloop)
return 0;
addtoopened(ntx);
//将ntx加入已扩展链表，并将这个节点从待扩展链表中去掉
flag = stretch(ntx);
//对ntx进行扩展,返回-1,0,1
if(flag == 1)
return 1;
}
}
int stretch(struct SPQ *ntx)
{
int fsr , fpr ; //在右岸上的人数
int fsl , fpl ; //在左岸上的人数
int
sst , spt ; //出发时在船上的人数
int ssr , spr ; //返回时船上的人数
struct SPQ *newnode;
for (sst = 0 ; sst <=
2 ; sst++) //讨论不同的可能性并判断是否符合条件
{
fsr = ntx -> sr;
fpr = ntx -> pr;
fsl = ntx -> sl;
fpl = ntx -> pl;
if ((sst <=
fsr) && (( 2 - sst) <=
fpr))//满足人数限制
{
spt = 2 - sst;
fsr = fsr - sst;
fpr = fpr - spt;
if((fpr ==
0) && (fsr ==
0))//搜索成功
{
newnode = (struct SPQ*) malloc (sizeof(spq));
if(newnode==NULL)
{
printf("\n内存不够！\n");
exit(0);
}
newnode -> upnode = ntx;
//保存父结点的地址以成链表
newnode -> nextnode = NULL;
newnode -> sr = 0;
newnode -> pr = 0;
newnode -> sl = opened -> sr;
newnode -> pl = opened -> pr;
newnode -> sst = sst;
newnode -> spt = spt;
newnode -> ssr = 0;
newnode -> spr = 0;
newnode -> loop = ntx -> loop + 1;
oend -> nextnode = newnode;
oend = newnode;
openednum++;
return 1;
}
else if ((fpr - fsr) * fpr >= 0) //判断是否满足传教士人数必须大于或等于野人人数
{
fsl = fsl + sst;
fpl = fpl + spt;
for (ssr = 0 ; ssr <= 1 ; ssr++)
//返回
{
int ffsl , ffpl;
if ((ssr <= fsl) && ((1 - ssr) <= fpl))
{
spr = 1 - ssr;
ffsl = fsl - ssr;
ffpl = fpl - spr;
if ((ffpl - ffsl) * ffpl >= 0)
{
//若符合条件则分配内存并付值
int
ffsr , ffpr;
ffsr = fsr + ssr;
ffpr = fpr + spr;
newnode = (struct SPQ*) malloc (sizeof(spq));
if(newnode==NULL)
{
printf("\n内存不够！\n");
exit(0);
}
newnode -> upnode = ntx;
//保存父结点的地址以成链表
newnode -> sr = ffsr;
newnode -> pr = ffpr;
newnode -> sl = ffsl;
newnode -> pl = ffpl;
newnode -> sst = sst;
newnode -> spt = spt;
newnode -> ssr = ssr;
newnode -> spr = spr;
newnode -> loop = ntx -> loop + 1;
uend -> nextnode = newnode;
uend = newnode;
unopenednum++;
}
}
}
}
}
}
return 0;
}
void addtoopened(struct SPQ *ntx)
{
unopened = unopened -> nextnode;
unopenednum--;
if (openednum == 0 )
oend = opened = ntx;
oend -> nextnode = ntx;
oend = ntx;
openednum++;
}
void recorder()
{
int i , loop;
struct SPQ *newnode;
struct SPQ *ntx;
loop = oend -> loop;
ntx = oend;
resultnum = 0;
for( i = 0 ; i <= loop ; i++ )
{
newnode = (struct SPQ*) malloc (sizeof(spq));
if(newnode==NULL)
{
printf("\n内存不够！\n");
exit(0);
}
newnode -> sr = ntx -> sr;
newnode -> pr = ntx -> pr;
newnode -> sl = ntx -> sl;
newnode -> pl = ntx -> pl;
newnode -> sst = ntx -> sst;
newnode -> spt = ntx -> spt;
newnode -> ssr = ntx -> ssr;
newnode -> spr = ntx -> spr;
newnode -> nextnode = NULL;
ntx = ntx -> upnode;
if(i == 0)
result = newnode;
newnode -> nextnode = result;
result = newnode;
resultnum++;
}
}
void releasemem()
{
int i;
struct SPQ* nodefree;
for ( i = 1 ; i < openednum ; i++ )
{
nodefree = opened;
opened = opened -> nextnode;
free(nodefree);
}
for ( i = 0 ; i < unopenednum ; i++ )
{
nodefree = unopened;
unopened = unopened -> nextnode;
free(nodefree);
}
}
void showresult()
{
int i;
int fsr , fpr ; //在右岸上的人数
int fsl , fpl ; //在左岸上的人数
struct SPQ* nodefree;
printf("%d个传教士",result -> pr);
printf("%d个野人",result -> sr);
printf("%d个传教士",result -> pl);
printf("%d个野人",result -> sl);
for ( i = 1 ; i < resultnum ; i++ )
{
nodefree = result;
result = result -> nextnode;
free(nodefree);
printf("\n\n\t左岸人数 船上人数及方向 右岸人数\n");
printf("第%d轮\n",i);
fpl = result -> pl - result -> spt + result -> spr;
fpr = result -> pr - result -> spr;
fsl = result -> sl - result -> sst + result -> ssr;
fsr = result -> sr - result -> ssr;
printf("传教士%8d%8d\t<-\t%8d\n",fpl,result -> spt,fpr);
printf("野
人%8d%8d\t<-\t%8d\n",fsl,result -> sst,fsr);
printf("传教士%8d%8d\t->\t%8d\n",result -> pl,result -> spr,result -> pr - result -> spr);
printf("野
人%8d%8d\t->\t%8d\n",result -> sl,result -> ssr,result -> sr - result -> ssr);
}
printf("\n全体传教士和野人全部到达对岸");
free(result);
}
void goon()
{
char choice;
for(;;)
{
printf("是否继续？(Y/N)\n");
scanf ("%s" , &choice);
choice=toupper(choice);
if(choice=='Y')break;
if(choice=='N')exit(0);
}
}
首先声明这个不是我写的，是我从网上下的源程序代码中的一个跟
这道题相似我就发出来大家来看一下，我还没有研究。哈哈
2009-12-21 18:51
