POJ 2488 A Knight's Journey --- DFS

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 32838		Accepted: 11182

Description


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? 

Problem 

Find a path such that the knight visits every square once. The knight can start and end on any square of the board.
Input
The 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, . . .
Output
The 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.
Sample Input

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

Source
TUD Programming Contest
2005, Darmstadt, Germany

#include <stdio.h>
#include <string.h>

int bhx[] = { -1, 1, -2, 2, -2, 2, -1, 1 };
int bhy[] = { -2, -2, -1, -1, 1, 1, 2, 2 };//注意顺序按字典序从小到大是这样的
bool vis[55][55];//标记数组
struct node
{
int x;
int y;
} ls[1010];//记录路径的队列

bool flag;//标记是否把所有点遍历

void dfs(int sx,int sy,int p,int q,int top)//dfs搜索遍历
{
if(flag)
return ;
if(top >= p * q)
{
flag = true;
for(int i = 0;i < top;i++)
{
printf("%c%d",(char)(ls[i].x + 'A' - 1),ls[i].y);//输出答案，这里不能在最后输出，因为top每次dfs都会变，这一次搜完了并不代表完全返回
}
printf("\n\n");
return ;
}
int a,b;
for(int i = 0; i < 8; i++)
{
a = sx + bhx[i];
b = sy + bhy[i];
if(a >= 1 && a <= p && b >= 1 && b <= q && !vis[a][b])
{
ls[top].x = b;
ls[top].y = a;
vis[a][b] = true;
dfs(a,b,p,q,top + 1);
vis[a][b] = false;
}
}
}

int main()
{
int t,p,q,c = 0,top;
scanf("%d",&t);
while(t--)
{
flag = false;
printf("Scenario #%d:\n",++c);
scanf("%d%d",&p,&q);
for(int i = 1; i <= q; i++)
{
for(int j = 1; j <= p; j++)
{
top = 0;
memset(vis,0,sizeof(vis));
ls[top].x = i;
ls[top].y = j;
top++;
vis[j][i] =true;
dfs(j,i,p,q,top);
if(flag)
break;
}
if(flag)
break;
}
top = p * q;
if(!flag)
printf("impossible\n\n");
//        else
//        {
//            for(int i = 0;i < top;i++)
//            {
//                printf("%c%d",(char)(ls[i].y + 'A' - 1),ls[i].x);
//            }
//            printf("\n\n");
//        }
}
return 0;
}