toj 1702 A Knight's Journey
Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?
ProblemFind a path such that the knight visits every square once. The knight can start and end on any square of the board.
InputThe input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 ≤ p * q ≤ 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .
OutputThe output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
Source: TUD Programming Contest 2005
Problem ID in problemset: 1702
// #include <stdio.h>
int t;
int main()
{
scanf( " %d " , & t);
int p,q,zz = 1 ;
while (t -- )
{
scanf( " %d%d " , & p, & q);
printf( " Scenario #%d:\n " ,zz ++ );
if (p == 1 && q == 1 )
printf( " A1\n " );
else if (p == 3 && q == 4 )
printf( " A1C2A3B1D2B3C1A2C3D1B2D3\n " );
else
if (p == 3 && q == 7 )
printf( " A1B3D2F1G3E2G1F3E1G2E3C2A3B1C3A2C1D3B2D1F2\n " );
else if (p == 3 && q == 8 )
printf( " A1B3C1A2C3D1B2D3E1G2E3C2A3B1D2F1H2F3G1E2G3H1F2H3\n " );
else
if (p == 4 && q == 3 )
printf( " A1B3C1A2B4C2A3B1C3A4B2C4\n " );
else
if (p == 4 && q == 5 )
printf( " A1B3C1A2B4D3E1C2D4E2C3A4B2D1E3C4A3B1D2E4\n " );
else
if (p == 4 && q == 6 )
printf( " A1B3C1A2B4C2D4E2F4D3E1F3D2B1A3C4B2A4C3E4F2D1E3F1\n " );
else
if (p == 5 && q == 4 )
printf( " A1B3A5C4D2B1A3B5D4C2B4A2C1D3C5A4B2D1C3D5\n " );
else
if (p == 5 && q == 5 )
printf( " A1B3A5C4A3B1D2E4C5A4B2D1C3B5D4E2C1A2B4D5E3C2E1D3E5\n " );
else
if (p == 6 && q == 4 )
printf( " A1B3A5C6D4B5D6C4D2B1A3C2B4A2C1D3B2D1C3D5B6A4C5A6\n " );
else
if (p == 7 && q == 3 )
printf( " A1B3C1A2C3B1A3C2B4A6C7B5A7C6A5B7C5A4B2C4B6\n " );
else
if (p == 8 && q == 3 )
printf( " A1B3C1A2B4C2A3B1C3A4B2C4A5B7C5A6B8C6A7B5C7A8B6C8\n " );
else
printf( " impossible\n " );
printf( " \n " );
}
return 0 ;
}
//比赛的时候自己错了一点,赛后交了是对的,唉,深搜一定要细心啊!!!!
// 不打表也是对的
#include < iostream >
#include < stack >
#define MAX 27
using namespace std;
bool used[MAX][MAX],mark;
int sum,p,q,t;
int a[ 8 ][ 2 ] = {{ - 1 , - 2 },{ 1 , - 2 },{ - 2 , - 1 },{ 2 , - 1 },{ - 2 , 1 },{ 2 , 1 },{ - 1 , 2 },{ 1 , 2 }};
int num;
typedef struct node
{
short num;
char ch;
node(){};
node( short nn, char ccc)
{
num = nn;
ch = ccc;
}
}Point;
stack < Point > S,RS;
Point temp;
void Init()
{
scanf( " %d%d " , & p, & q);
int i,j;
for (i = 0 ;i < q;i ++ )
for (j = 0 ;j < p;j ++ )
used[i][j] = false ;
}
bool Bound( int x, int y)
{
if (x >= 0 && y >= 0 && x < p && y < q)
return true ;
else
return false ;
}
void DFS( int x, int y, int sum) // x是列,y是行
{
int i,j,tx,ty;
if (sum == p * q)
{
mark = true ;
return ;
}
for (i = 0 ;i < 8 ;i ++ )
{
tx = x + a[i][ 1 ]; // tx是列
ty = y + a[i][ 0 ]; // ty是行
if ( ! Bound(ty,tx) || used[tx][ty])
{
continue ;
}
used[tx][ty] = true ;
S.push(node(ty,( char )( ' A ' + tx)));
sum += 1 ;
DFS(tx,ty,sum);
if ( ! mark)
S.pop();
if (mark)
return ;
used[tx][ty] = false ;
sum -= 1 ;
}
}
int main()
{
int i,j;
int zz = 1 ;
scanf( " %d " , & t);
while (t -- )
{
Init();
printf( " Scenario #%d:\n " ,zz ++ );
mark = false ;
num = 0 ;
for (i = 0 ;i < q;i ++ )
{
if (mark)
break ;
for (j = 0 ;j < p;j ++ )
{
used[i][j] = true ;
while ( ! S.empty())
S.pop();
sum = 1 ;
temp.ch = ( char )( ' A ' + i);
temp.num = j;
S.push(temp);
DFS(i,j,sum); // i是列,j是行
if (mark)
break ;
used[i][j] = false ;
}
}
if ( ! mark)
printf( " impossible\n\n " );
else
{
while ( ! S.empty())
{
temp = S.top();
S.pop();
RS.push(temp);
}
while ( ! RS.empty())
{
temp = RS.top();
RS.pop();
printf( " %c%d " ,temp.ch,temp.num + 1 );
}
printf( " \n\n " );
}
}
return 0 ;
}