Data Structure 0103 Nottingham & Eight Queens

The recursive algorithm is mainly realized from Hanoi and eight Queens problem.

Hanoi

#include <stdio.h>

void Move (int n, char X,char y, char z)
{
if (1==n)
{
printf ("%c-->%c\n", x,z);
}
Else
{
Move (N-1,x,z,y); Move n-1 plates from X to Y with Z
printf ("%c-->%c\n", x,z); Move the nth plate from X to Z
Move (N-1,Y,X,Z); Move n-1 plates from Y to Z with x
}
}

int main ()
{
int n;

printf ("Please enter the number of layers Hanoi: \ n");
scanf ("%d", &n);
printf ("Move the steps below: \ n");
Move (n, ' X ', ' Y ', ' Z ');

return 0;
}

Eight queens:

Put eight queens on the 8*8 chess, so that they cannot attack each other, that is, any two queens cannot be in the same row, the same column, ask how many kinds of pendulum (92 kinds, poor lifting method):

The code is as follows:

#include <stdio.h>
#include <stdlib.h>

int count=0;

int notdanger (int row,int j,int (*chess) [8])
{
int i,k,flag1=0,flag2=0,flag3=0,flag4=0,flag5=0;
//Judging column orientation
for (i=0;i<8;i++)
{
if (* (* (chess+i) +j)!=0)
{
flag1=1;
break;
}
}
//Judging upper left
for (i=row,k=j;i>=0 && k>=0;i--, k--)
{
if (* (* (chess+i) +k)!=0)
{
Flag2=1;
break;
}
}
//judgment right bottom
for (i=row,k=j;i<8 && k<8;i++,k++)
{
if (* (* (chess+i) +k)!=0)
{
Flag3=1;
break;
}
}
//judgment upper right
for (i=row,k=j;i>=0 && k<8;i--, k++)
{
if (* (* (chess+i) +k)!=0)
{
Flag4=1;
break;
}
}
//Judging lower left
for (i=row,k=j;i<8 && k>=0;i++,k--)
{
if (* (* (chess+i) +k)!=0)
{
Flag5=1;
break;
}
}
if (Flag1 | | flag2 | | flag3 | | flag4 | | flag5)
{
return 0;
}
Else
{
return 1;
}
}

void Eightqueen (int row, int n, int (*chess) [8])//Three parameters represent rows, columns, and pointers to expected each row
{
int i,j,chess2[8][8];

for (i=0;i<8;i++)
{
for (j=0;j<8;j++)
{
CHESS2[I][J]=CHESS[I][J];
}
}

if (8==row)
{
printf ("%d \ n", count+1);
for (i=0;i<8;i++)
{
for (j=0;j<8;j++)
{
printf ("%d", * (* (chess2+i) +j));
}
printf ("\ n");
}
printf ("\ n");
count++;
}
Else
{
for (j=0;j<n;j++)
{
if (Notdanger (row,j,chess))//Determine if there are two queen conflicts in this position
{
for (i=0;i<8;i++)
{
* (* (chess2+row) +i) = 0;
}
* (* (chess2+row) +j) = 1;
Eightqueen (ROW+1,N,CHESS2);
}
}
}
}

int main ()
{
int chess[8][8],i,j;

for (i=0;i<8;i++)//0 means no data, 1 means to store queen
{
for (j=0;j<8;j++)
{
chess[i][j]=0;
}
}

Eightqueen (0, 8, chess);

printf ("There are a total of%d workarounds!") \ n ", count);

return 0;
}

Data Structure 0103 Nottingham & Eight Queens