华容道每一关的全过程

华容道已经被研究过多年,也总结了许多关口的走法,为让各位喜欢华容道的朋友少走弯路,我把一些走法整理出来,与大家分享。

下面的走法沿用L.E.Hordern的记录方法,即在多数情况下只要指明走哪一个棋子就够了,只有少数情况下才指明如何走。这时用以下符号来表示。L向左;R向右;U向上;D向下;!只走一格;#必须拐弯(指最小棋子)。没有这些符号时,表示直走,到头为止(一格或两格)。棋子编号见图1。当然,这只是指出了如何过关,大家也不必死记硬背这些步骤,关键要从此研究出过关的必要条件,而达到通关的目的。

(1) 横竖皆将

6 4 5 7 # 9 6 8 3 5 7 9 L 2 A 7 5 1 7 L A 2 4 5 9 L 4 5 8#3 1 9 L 4 5 8#3 1 9 L 4 5# 2A 9 # 4 1 3 6 8 5 2 A 9 7 4 3 5 8 6 D 3 A 9 1 7 4 3 1 2 2 6R 5# 8# A 9 1 7 4 3 1 A 9 1 7 2 6 8 5 A 9 3 4 2 6 5 # A

(2)守口如瓶之一

5 7L 2 A 1 3 6 4 1 A 2 7# 9 8 4 1 6 #4 1 6 5 #7 9 5 6 #1 4 7 # 9 5#2 A 7 #9 4 1 8 6 D 5 2 A 7 3 9 1 5 6 7 1 4 D 1 A 7 1 3 9 1 4 2 8 R 5 #6#A 7 1 3 9 1 4 A 8 3 2 8 6 5 A 7 1 9 2 8 5#A

(3)守口如瓶之二

7#9 8 6 #3 1 A 2 4 7 R 2 A 1 3 6 #8 9 7#4 A 5 6 #8 9 7 # 8 9 3 6# 51 6 U 5 1 A 4 81 2U 8 1 1 7 9 3 5 2#8 7 # 4 A 2#8 5 3 9 1 7 4 A 2 6 8 3 7 1 9 5 D 3 9 2 1 6 8 3 5 4 9 R 1# 7# A 2 1 6 8 3 5 A 2 1 6 4 A 7 1 A 2 3 8 4 9 1#A

(4)层层设防之二

9 L8#4 2 A 1 3 5 2 4 8 9 6 7 2 5 3 1 L,A 4 5 2 7 6 9 8 2 7 6 # 7 8# 7 9 3 6 # 5 8 #4 A 6# 5 3 8 9 2 4 A 6 1 5 8# A 6 1 1 5 8 3 4 7 2U 9 7 2 A 6 1# 4 A 6 3 2 6# 7 9 A 1#3 2 8 5 3 1 A 9 7 1# A 4 3 2 # A 1 6# 8 A 1 4 3 1# 4 3 9 7 8 6 D A 6 2 1 4 3 9 7 6 8 A 9 7 8 #A

(5)Top secret

7 5 3 2 1 4 6 7 L A 1#4 6 7 1 1 3 5 9 8 A 1 4 2 5 3# 4 7 R 6 2 4 1 A 8 9 3 D 5 1 4 2 7 U 6 U A 1 3 9 8 3 D 1 D A 7D 6D 2 5 4 9 8 3 1 A 9 8 1#A

(6)三军联防

6 7 4 3 7# 3 4 2 1 A 7 5 8 4 6 9# 6 4 8 3 9 L 2 1 A 5# 3 8 9 U 4 6 2 1 A5 7

3 9# A 1 2 4 6 8 9 A 1 2 4 6 9# A 3 7 5 1 2 4 6 9 8 A 4 6 8#A

(7)堵塞要道

5 9 6 7 4#2 A 3 #7 5 6 9 8 4 2 D A 3 1 7 5 6 9 8 4 2 D A 1 3 D 7 5 6 9 8 4 2 A 9 8 2#A

(8)水泄不通

9 7 6 8 9 U 7 6 5 4 8 9 U 5 4 9 A 1 3# 8 A 1 2 9 1# 4 5 A 3# 21# 4 5 6 7 A 5 4 1# 2 3 #5 4 2 1 9 D 3 8 5 4 A 7 6 1# 9 3 8#5 4 A 1 9 6 7 1 9 D A 4 5 2 8 3 U 6 7 9 1 A 6 7 1#A

(9)四路进兵(原文 67步,11 66步)

A 4 3 #2 A 4 3 #1 5 2 #7 6 A 3 #1 2 #7 6 9 8 A 6 7 2 0#1 3 #6 7 1 2 5 D 3 4 6 7 A 8 9 2# 5 3 4# 6 7 A 2 5 9 8 2 5 D A 7 6 1 4 3 U 9 8 5 2 A 9 8 2# A

华容道问题用计算机求解,一般采用广度搜索的方法,其原理很简单,就是把下一步可能有的走法全部算出来,比如第一步有五种走法,将这五种走法的下一步走法分别算出来,可能会有三十步,在继续将这三十步走法的下一步走法分别算出来,可能会更多,以此类推,直到达到目标状态(曹操在出口位置)为止。

在解华容道的问题上,我觉得有两个问题比较棘手。

其一、算法的效率。

其二、获得最优解法。

我是这样解决的:

1、 要提高算法的效率,首先要知道算法的瓶颈在什么地方,在得出每一个状态(走完一步各个棋子的位置)都要和前面的状态进行比较,以保证不重复,随着步数的增多,状态数会大幅度增加,这是,和前面的状态比较这一过程成了整个算法的效率。解决的办法,从两个地方着手,其一,增加每一步比较的速度。在程序中,用5*4的数组表示一个状态,这样,每一次比较要比较二十个数,因为数组中每个数定义从0-7,用三个二进制位可以表示,3*20=60位,用一个64位数就可以表示(有的资料说用四个字节就可以,我实在想不出来),这样每次比较一个64位数就可以了。其二、减少比较的状态,这是提高效率的关键。比较的时候不要和前面所有的状态都进行比较,只要和前两步的所有状态进行比较就可以了。经过以上的优化,在解横刀立马时,大约需要一,两秒钟就可以了,(我的机器,赛扬1.1OC1.46)。

2、 获得最优解法,比如横刀立马是81步,这里的一步指移动一个棋子,可以把一个卒子向一个方向移动两格,或者卒子拐弯移动两格,或者一个将向一个方向移动两格(横将横着移,竖将竖着移)都是一步。获得最优解法的关键是把下一步可能有的走法全部算出来,不能遗漏。我是根据空格来算走法的的,分三种情况:

① 、卒子拐弯移动,如果有连着两个空格(横向的),则如果在它的上面或下面(有四个位置)有卒子的话,那么可以拐弯移动,有四种走法。如果两个空格是竖向的,那么,空格的左右如果有卒子,也可以拐弯移动,也有四种走法。

②、向一个方向移动两格,这里可能出现的情况有:卒子向一个方向移动两格,横将横着移两格,竖将竖着移两格

③、考虑向一个方向移动一格的情况,这里情况很多,我不一一列举了。

以上的算法很麻烦,很大一部分程序用来写这个了,如果大家有更简单的,可以告诉我,但一个原则,必须把所有的走法全部考虑。

另外,说一下我在写程序时的小插曲。程序快写好时,运行时发现,每解一次,内存使用会增加7,8兆,后来发现分配的内存每释放导致的,其实在函数中也就分配了几十个字节,由于被重复调用,最后泄漏的内存就很可观了,以后使用指针分配内存可要注意了,(C用malloc,C++用new),一定要释放,弄不好,^@^。

程序用dev-C++ 4.9.9.0(可以从网上下,只有十多兆)编译通过,因为dev C++没有框架等东西,所以界面直接用window API写的。生成的可执行文件很小,68 K。另外,在程序中可以自定义布局,用5*4数表示。其中0-空格,1-卒子,2到6 将,7曹操。

最后附上所有的源代码。

main.cpp程序为:

#include <string>

#include <windows.h>

#include "HRD_Calculate.h"

char str[80];

PAINTSTRUCT pa;

HDC hdc,memdc;

RECT rect;

HBITMAP hbit;

HBRUSH hbrush;

HPEN hpen;

POINT point;

hrd_calculate hrd; // User declarations

int current_step;

unsigned __int8 display_node[5][4];

/* Declare Windows procedure */

LRESULT CALLBACK WindowProcedure (HWND, UINT, WPARAM, LPARAM);

/* Make the class name into a global variable */

char szClassName[ ] = "WindowsApp";

int WINAPI WinMain (HINSTANCE hThisInstance,

HINSTANCE hPrevInstance,

LPSTR lpszArgument,

int nFunsterStil)

{

HWND hwnd; /* This is the handle for our window */

MSG messages; /* Here messages to the application are saved */

WNDCLASSEX wincl; /* Data structure for the windowclass */

/* The Window structure */

wincl.hInstance = hThisInstance;

wincl.lpszClassName = szClassName;

wincl.lpfnWndProc = WindowProcedure; /* This function is called by windows */

wincl.style = CS_DBLCLKS; /* Catch double-clicks */

wincl.cbSize = sizeof (WNDCLASSEX);

/* Use default icon and mouse-pointer */

wincl.hIcon = LoadIcon (NULL, IDI_APPLICATION);

wincl.hIconSm = LoadIcon (NULL, IDI_WINLOGO);

wincl.hCursor = LoadCursor (NULL, IDC_ARROW);

wincl.lpszMenuName = NULL; /* No menu */

wincl.cbClsExtra = 0; /* No extra bytes after the window class */

wincl.cbWndExtra = 0; /* structure or the window instance */

/* Use Windows's default color as the background of the window */

wincl.hbrBackground = (HBRUSH) COLOR_BTNFACE;

/* Register the window class, and if it fails quit the program */

if (!RegisterClassEx (&wincl))

return 0;

/* The class is registered, let's create the program*/

hwnd = CreateWindowEx (

0, /* Extended possibilites for variation */

szClassName, /* Classname */

"华容道", /* Title Text */

WS_OVERLAPPED|WS_CAPTION|WS_SYSMENU, /* default window */

CW_USEDEFAULT, /* Windows decides the position */

CW_USEDEFAULT, /* where the window ends up on the screen */

544, /* The programs width */

375, /* and height in pixels */

HWND_DESKTOP, /* The window is a child-window to desktop */

NULL, /* No menu */

hThisInstance, /* Program Instance handler */

NULL /* No Window Creation data */

);

/* Make the window visible on the screen */

ShowWindow (hwnd, nFunsterStil);

/* Run the message loop. It will run until GetMessage() returns 0 */

while (GetMessage (&messages, NULL, 0, 0))

{

/* Translate virtual-key messages into character messages */

TranslateMessage(&messages);

/* Send message to WindowProcedure */

DispatchMessage(&messages);

}

/* The program return-value is 0 - The value that PostQuitMessage() gave */

return messages.wParam;

}

/* This function is called by the Windows function DispatchMessage() */

LRESULT CALLBACK WindowProcedure (HWND hwnd, UINT message, WPARAM wParam, LPARAM lParam)

{

int initx=20,inity=20,grid=50,interspace=3,arc=25;

int i,j,m=0;

char s[100];

switch (message) /* handle the messages */

{

case WM_CREATE:

{

CreateWindow("BUTTON","解题",WS_CHILD|WS_VISIBLE|BS_PUSHBUTTON,350,150,100,

30,hwnd,(HMENU)1000,((LPCREATESTRUCT) lParam)->hInstance,NULL);

CreateWindow("BUTTON","自定义布局",WS_CHILD|WS_VISIBLE|BS_PUSHBUTTON,350,90,100,

30,hwnd,(HMENU)1001,((LPCREATESTRUCT) lParam)->hInstance,NULL);

CreateWindow("EDIT","27732773144115510660",WS_CHILD|WS_VISIBLE|ES_NUMBER|WS_BORDER,350,50,165,

20,hwnd,(HMENU)1002,((LPCREATESTRUCT) lParam)->hInstance,NULL);

GetClientRect(hwnd,&rect);

hdc=GetDC(hwnd);

memdc=CreateCompatibleDC(hdc);

hbit=CreateCompatibleBitmap(hdc,rect.right,rect.bottom);

SelectObject(memdc,hbit);

hbrush = (HBRUSH) GetStockObject(WHITE_BRUSH);

SelectObject(memdc, hbrush);

//hpen = (HPEN) GetStockObject(BLACK_PEN);

//SelectObject(memdc, hpen);

ReleaseDC(hwnd,hdc);

///////////////////////////////////////

display_node[0][0]=GENERAL1;

display_node[0][1]=CAOCAO;

display_node[0][2]=CAOCAO;

display_node[0][3]=GENERAL2;

display_node[1][0]=GENERAL1;

display_node[1][1]=CAOCAO;

display_node[1][2]=CAOCAO;

display_node[1][3]=GENERAL2;

display_node[2][0]=GENERAL3;

display_node[2][1]=GENERAL5;

display_node[2][2]=GENERAL5;

display_node[2][3]=GENERAL4;

display_node[3][0]=GENERAL3;

display_node[3][1]=SOLDIER;

display_node[3][2]=SOLDIER;

display_node[3][3]=GENERAL4;

display_node[4][0]=SOLDIER;

display_node[4][1]=BLANK;

display_node[4][2]=BLANK;

display_node[4][3]=SOLDIER;

break;

}

case WM_TIMER:

{

if(current_step<hrd.depth)

current_step++;

else

{

current_step=0;

KillTimer(hwnd,1);

Sleep(2000);

}

for( i=0;i<5;i++)

for( j=0;j<4;j++)

display_node[i][j]=hrd.out[current_step].state[i][j];

InvalidateRect(hwnd, NULL, 0);

break;

}

case WM_COMMAND:

if(HIWORD(wParam)==BN_CLICKED)

switch (LOWORD(wParam))

{

case 1000:

{

//hrd= new hrd_Calculate;

hrd.InitState(display_node);

if( hrd.SearchNode())

{

sprintf(s, "解题成功!\n\n解题深度:%d 节点数:%d", hrd.depth,hrd.totalnodes);

MessageBox(hwnd,s,"华容道",MB_OK);

hrd.OutputStep();

current_step=0;

SetTimer(hwnd, 1,700, NULL);

}

else

{

sprintf(s,"此局无解") ;

MessageBox(hwnd,s,"华容道",MB_OK);

}

break;

}

case 1001:

{

GetDlgItemText(hwnd,1002,str,80);

for (i=0;i<5;i++)

for(j=0;j<4;j++)

{

display_node[i][j]=(int)(str[m])-0x30;

m++;

}

InvalidateRect(hwnd, NULL, 1);

break;

}

}

break;

case WM_PAINT:

{

hdc = BeginPaint(hwnd,&pa);

PatBlt(memdc, 0, 0, rect.right, rect.bottom, PATCOPY);

//Draw

for (i=0;i<5;i++)

for(j=0;j<4;j++)

{

if (display_node[i][j]==SOLDIER)

RoundRect(memdc,inity+j*grid+j*interspace,initx+i*grid+i*interspace,

inity+(j+1)*grid+j*interspace,initx+(i+1)*grid+i*interspace,arc,arc);

if (display_node[i][j]>=GENERAL1 && display_node[i][j]<=GENERAL5)

{

if (i<4)

if (display_node[i][j]==display_node[i+1][j])

RoundRect(memdc,inity+j*grid+j*interspace,initx+i*grid+i*interspace,

inity+(j+1)*grid+j*interspace,initx+(i+2)*grid+(i+1)*interspace,arc,arc);

if (j<3)

if (display_node[i][j]==display_node[i][j+1])

RoundRect(memdc,inity+j*grid+j*interspace,initx+i*grid+i*interspace,

inity+(j+2)*grid+(j+1)*interspace,initx+(i+1)*grid+i*interspace,arc,arc);

}

if (display_node[i][j]==CAOCAO)

if (i<4 && j<3)

if( display_node[i+1][j+1]==CAOCAO)

RoundRect(memdc,inity+j*grid+j*interspace,initx+i*grid+i*interspace,

inity+(j+2)*grid+(j+1)*interspace,initx+(i+2)*grid+(i+1)*interspace,arc,arc);

}

//////////////////////////////////

BitBlt(hdc,0,0,rect.right,rect.bottom,memdc,0,0,SRCCOPY);

EndPaint(hwnd,&pa);

break;

}

case WM_DESTROY:

{

PostQuitMessage (0); /* send a WM_QUIT to the message queue */

DeleteDC(memdc);

DeleteObject(hbit);

break;

}

default: /* for messages that we don't deal with */

return DefWindowProc (hwnd, message, wParam, lParam);

}

return 0;

}

///HRD_Calculate.h 的程序写法

/////////////////////////////////////////////////

//华容道解法1.0.0.1

//此解法可得出最优解

//横刀立马 81步

//最后修改时间 2004.9.22 晚上

//

/////////////////////////////////////////////////

#include "HRD_Calculate.h"

hrd_calculate::hrd_calculate()

{

//申请状态表空间

first= new s_node[MAX_NODES];

}

hrd_calculate::~hrd_calculate()

{

delete[] first;

}

void hrd_calculate::NodeToSNode(node * pnode,s_node* psnode)

{

int i,j;

__int8 hgeneral=8,vgeneral=9;

node * tnode= new node;

*tnode=*pnode;

for( i=0;i<5;i++)

for( j=0;j<4;j++)

{

if (tnode->state[i][j]>=GENERAL1 && tnode->state[i][j]<=GENERAL5)

{

if (j<3)

if (tnode->state[i][j] == tnode->state[i][j+1])

{

tnode->state[i][j]=hgeneral;

tnode->state[i][j+1]=hgeneral;

}

if(i<4)

if(tnode->state[i][j] == tnode->state[i+1][j])

{

tnode->state[i][j]=vgeneral;

tnode->state[i+1][j]=vgeneral;

}

}

}

for( i=0;i<5;i++)

for( j=0;j<4;j++)

{

if(tnode->state[i][j]==hgeneral) tnode->state[i][j]=HGENERAL;

if(tnode->state[i][j]==vgeneral) tnode->state[i][j]=VGENERAL;

}

psnode->prior=(s_node *)pnode->prior;

psnode->state=0;

psnode->ext_state=0;

for( i=0;i<5;i++)

for( j=0;j<4;j++)

{

psnode->state += pnode->state[i][j];

psnode->ext_state += tnode->state[i][j];

if (!(i==4 && j==3)) psnode->state = psnode->state<<3;

if (!(i==4 && j==3)) psnode->ext_state = psnode->ext_state<<3;

}

delete tnode;

}

void hrd_calculate::SNodeToNode(s_node* psnode,node * pnode)

{

__int64 temp,s;

s = psnode->state;

pnode->prior=(node*)psnode->prior;

for(int i=4;i>=0;i--)

for(int j=3;j>=0;j--)

{

temp = s & 0x0000000000000007;

pnode->state[i][j]= temp ;

s = s >>3 ;

}

}

void hrd_calculate::OutputStep()

{

node * outfirst,* outlast,*p;

outfirst=&out[0];

outlast=outfirst+(depth);

p=outlast;

while ( p>=outfirst)

{

SNodeToNode(last,p);

last=last->prior;

p--;

};

}

bool hrd_calculate::SearchNode()

{

int nextnodes;

node * tnode=new node;

int total;

while(true)

{

nextnodes=0;

table[depth+1]=(unsigned int)(last+1);

for ( ;search<=current_last ; search++)

{

SNodeToNode(search,tnode);

tnode->prior=(node *)search;

total=SearchOneNode(tnode);

nextnodes +=total;

if (total==SUCCESS)

{

delete tnode;

return true;

}

}

if (nextnodes==0)

{

delete tnode;

return false;

}

depth++;

current_last=last;

}

}

int hrd_calculate::AddNode(node c)

{

s_node *p;

s_node *snode=new s_node;

if (depth<=3) p=first;

else p=(s_node*)table[depth-1];

NodeToSNode(&c,snode);

for (;p<=last;p++)

if (p->ext_state== snode->ext_state)

{

delete snode;

return ADD_NO_NODE;

}

//加入节点

last++;

last->prior=snode->prior;

last->state=snode->state;

last->ext_state=snode->ext_state;

totalnodes++;

delete snode;

if (c.state[3][1]==CAOCAO && c.state[4][2]==CAOCAO )

return SUCCESS;

else

return ADD_ONE_NODE;

}

void hrd_calculate::InitState(unsigned __int8 state[5][4])

{

//设定初始状态

node initnode;

initnode.prior=0; //没有上一步

for(int i=0;i<5;i++)

for(int j=0;j<4;j++)

initnode.state[i][j]=state[i][j];

////////////////////

NodeToSNode(&initnode,first);

////////////

last=first;

search=first;

current_last=first;

depth=1;

totalnodes=1;

table[0]=0;

table[depth]=(unsigned int)first;

}

int hrd_calculate::SearchOneNode(node *c)

{

int i,j;

int next_nodes=0;

node t;

for(i=0;i<5;i++)

for(j=0;j<4;j++)

{

if (c->state[i][j]==BLANK)

{

///////////////////////////////////////////////////////////////////////////////

//直走两步

if (j<3)

{

if (c->state[i][j+1]==BLANK)

{

if (j>0)//左边兵右移两格

{

if (c->state[i][j-1] == SOLDIER)

{

t=*c; t.prior=c->prior;

t.state[i][j-1]=BLANK;

t.state[i][j+1]=SOLDIER;

switch (AddNode(t))

{

case SUCCESS: return SUCCESS;

case ADD_ONE_NODE: next_nodes++;

}

}

}

if (j<2)//右边兵左移两格

{

if (c->state[i][j+2]==SOLDIER)

{

t=*c; t.prior=c->prior;

t.state[i][j+2]=BLANK;

t.state[i][j]=SOLDIER;

switch (AddNode(t))

{

case SUCCESS: return SUCCESS;

case ADD_ONE_NODE: next_nodes++;

}

}

}

if (j==2)//左边将右移两格

{

if (c->state[i][j-1]>=GENERAL1 && c->state[i][j-1]<=GENERAL5 && c->state[i][j-1]==c->state[i][j-2])

{

t=*c; t.prior=c->prior;

t.state[i][j]=c->state[i][j-1];

t.state[i][j+1]=c->state[i][j-1];

t.state[i][j-1]=BLANK;

t.state[i][j-2]=BLANK;

switch (AddNode(t))

{

case SUCCESS: return SUCCESS;

case ADD_ONE_NODE: next_nodes++;

}

}

}

if (j==0)//右边将左移两格

{

if (c->state[i][j+2]>=GENERAL1 && c->state[i][j+2]<=GENERAL5 && c->state[i][j+2]==c->state[i][j+3])

{

t=*c; t.prior=c->prior;

t.state[i][j]=c->state[i][j+2];

t.state[i][j+1]=c->state[i][j+2];

t.state[i][j+2]=BLANK;

t.state[i][j+3]=BLANK;

switch (AddNode(t))

{

case SUCCESS: return SUCCESS;

case ADD_ONE_NODE: n