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背景：拼图游戏！
要将以下的一数组
1 3 0
4
6 2 7
8
5 9 10 12
13 14 11 15
变成以下数组
1 2
3 4
5 6 7
8
9 10 11 12
13 14 15 0
0
代表空
最少需要多少步吗？该怎么走呢？有几种走法？
以下是我编的c
语言：
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#define SNUM 100
int a[SNUM][SNUM],line,max,NUM;
int c[]={1,0,-1,0},d[]={0,-1,0,1};
int ii[100],jj[100],g,finish=0;
int iii[100],jjj[100];
/**********************
从文件中提取数组****************************/
void geta(int (*p)[SNUM])
{FILE *fp;
int b[100],i,j,k=0;
if((fp=fopen("D://in.txt","r"))==NULL)
{printf("can't not open the file/n");
exit(0);
}
while(fscanf(fp,"%d",&b[k])!=EOF)
{
k++;
}
line=sqrt(k);
k=0;
for(i=0;i<line;i++)
for(j=0;j<line;j++)
{
p[i][j]=b[k];
k++;
}
fclose(fp);
}
/***************
判断数组是否满足要求****************************/
int ifright(int (*p)[SNUM])
{ int i,j,n=1,m=1;
for(i=0;i<line;i++)
for(j=0;j<line;j++)
{if(p[i][j]==n)
m++;
n++;
}
if(n==m+1&&p[line-1][line-1]==0)
return 1;
else
return 0;
}
/****************
获取0
的行*********************************/
int geti(int (*p)[SNUM])
{int i,j,n;
for(i=0;i<line;i++)
for(j=0;j<line;j++)
if(p[i][j]==0)
n=i;
return n;
}
/****************
获取0
的列***********************************/
int getj(int (*p)[SNUM])
{int i,j,n;
for(i=0;i<line;i++)
for(j=0;j<line;j++)
if(p[i][j]==0)
n=j;
return n;
}
/*****************
向文件中输入二维数组***********************/
void print(int (*p)[SNUM])
{int i,j;
char a='/n';
FILE *fp1;
if((fp1=fopen("D://out.txt","a"))==NULL)
{printf("can't create the file/n");
exit(0);
}
for(i=0;i<line;i++)
{ for(j=0;j<line;j++)
fprintf(fp1,"%5d",p[i][j]);
fputc(a,fp1);
fputc(a,fp1);
}
fclose(fp1);
}
/*****************
向文件中输入一维数组*************************/
void printa(int *p)
{int i;
char a='/n';
FILE *fp2;
if((fp2=fopen("D://out.txt","a"))==NULL)
{printf("can't create the file/n");
exit(0);
}
for(i=0;i<max-1;i++)
fprintf(fp2,"%5d",p[i]);
fputc(a,fp2);
fclose(fp2);
}
/*****************
向文件中输入一个数*************************/
void printc(int n)
{FILE *fp3;
char a='/n';
if((fp3=fopen("D://out.txt","a"))==NULL)
{printf("can't create the file/n");
exit(0);
}
fprintf(fp3,"%d",n);
fputc(a,fp3);
fclose(fp3);
}
/*****************
向文件中输入一字符串***********************/
void printx(char *p)
{FILE *fp3;
char a='/n';
if((fp3=fopen("D://out.txt","a"))==NULL)
{printf("can't create the file/n");
exit(0);
}
fputs(p,fp3);
fputc('/n',fp3);
fputc('/n',fp3);
fclose(fp3);
}
/*****************
使文件内容清空**********************************/
void clear()
{FILE *fp3;
if((fp3=fopen("D://out.txt","w"))==NULL)
{printf("can't create the file/n");
exit(0);
}
fclose(fp3);
}
/****************
递归函数（主体部分，用了回溯法）******************/
int tray(int (*p)[SNUM],int i,int j,int li,int lj,int num,int max)
{ int q=0,v,u,k,temp,kk;
if(num<max)
{ if(ifright(p))
{ print(p);
printx("
要移动的最小步数：");
printc(num);
printx("
坐标:");
printa(ii);
printa(jj);
for(kk=0;kk<num;kk++)
{iii[kk]=ii[kk];
jjj[kk]=jj[kk];
}
NUM=num;
q=1;
finish=1;
}
for(k=0;k<4;k++)
{u=i+c[k];
v=j+d[k];
if(u>=0&&u<line&&v>=0&&v<line)
{ if(u==li&&v==lj)
continue;
else
{p[i][j]=p[u][v];
temp=p[u][v];
p[u][v]=0;
ii[num]=u;
jj[num]=v;
q=tray(p,u,v,i,j,num+1,max);
if(q=1)
{p[u][v]=temp;
p[i][j]=0;
}
}
}
}
}else
q=1;
return q;
}
/****************
主函数**********************************/
void main()
{int xi,xj,i,j,k=0;
geta(a);
xi=geti(a);
xj=getj(a);
clear();
printx("the start array is:");
print(a);
printx("the finish array is:");
for(max=0;max<1000;max++)
{ tray(a,xi,xj,xi,xj,0,max);
if(finish==1)
break;
}
printx("the game start!!");
while(k<NUM)
{i=iii[k];
j=jjj[k];
a[xi][xj]=a[i][j];
xi=iii[k];
xj=jjj[k];
a[i][j]=0;
k++;
printx("next:");
print(a);
}
printx("the game over!!");
}
以上是原代码（总共是198
行）
结果：（内容在out.txt
中）
the start array is:
1 3 0
4
6 2 7
8
5 9 10 12
13 14 11 15
the finish array is:
1 2 3
4
5 6 7
8
9 10 11 12
13 14 15 0
要移动的最小步数：
8
坐标:
0 1 1
2 2 2 3
3
1 1 0
0 1 2 2
3
the game start!!
next:
1 0 3
4
6 2 7
8
5 9 10 12
13 14 11 15
next:
1 2 3
4
6 0 7
8
5 9 10 12
13 14 11 15
next:
1 2 3
4
0 6 7
8
5 9 10 12
13 14 11 15
next:
1 2 3
4
5 6 7
8
0 9 10 12
13 14 11 15
next:
1 2 3
4
5 6 7
8
9 0 10 12
13 14 11 15
next:
1 2 3
4
5 6 7
8
9 10 0 12
13 14 11 15
next:
1 2 3
4
5 6 7
8
9 10 11 12
13 14 0 15
next:
1 2 3
4
5 6 7
8
9 10 11 12
13 14 15 0
the game over!!