poj 2488 A Knight's Journey 【dfs】【字典序】【刷题计划】

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 47516		Accepted: 16161

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

题意：在一个n*m的国际棋盘上，象棋以马走日的方式走，行数为1-n的数字,列数为A-A+m的字母，按照字典序输出一条合适的路径。

思路：先构造一个棋盘数组和一个标记数组，按照字典序的方式就是先走字典序优先的位置，所以在定义方向数组的时候需要留意方位顺序，最后，每个样例后都一个空行。

喵？喵？喵？这道题的字典序操作真是毁我青春，我一直以为字典序就是先按照字母表顺序输出一遍再从头开始，然后就躺尸在这道题上，今天师父一看我屏幕，感叹这种水题居然还没有写过（毕竟上个周我就缠着师父给我讲为啥错了），最后师父才发现，原来我徒弟连字典序都不会（呜呜呜，谁知道是这样的字典序啊，样例误导人~），师父讲完字典序以后我就A过啦~

#include<stdio.h>
#include<string.h>
#define N 100
int map

,m,n,flag;
char str

;
struct node1{
int x;
char y;
}l[N*N];

struct node2{
int x;
char y;
}f[N*N];

void dfs(int x,int y,int step)
{
int k[8][2] = {-1,-2,1,-2,-2,-1,2,-1,-2,1,2,1,-1,2,1,2};
int nx,ny,i;
if(step > n*m)//在找到合适路径的情况下
{
flag = 1;
for(i = 1; i <= n*m; i ++)
{
f[i].x = l[i].x ;//将临时数组的值存入最终数组
f[i].y = l[i].y ;
}
return;
}
else
{
for(i = 0; i < 8; i ++)
{
if(!flag)//在未找到合适路径的情况下
{
nx = x + k[i][0];
ny = y + k[i][1];
if(nx < 1||ny < 1||nx > n||ny > m||map[nx][ny] == 1)
continue;
map[nx][ny] = 1;//标记为已经访问过
l[step].x = nx;//用临时数组存储下标
l[step].y = str[nx][ny];
dfs(nx,ny,step +1);//往下搜索
map[nx][ny] = 0;//回溯
}

}
}
}
int main()
{
int i,j;
int t,t2=0;
scanf("%d",&t);
while(t --)
{
scanf("%d%d",&n,&m);
flag = 0;
memset(map,0,sizeof(map));//
memset(l,0,sizeof(l));//
memset(f,0,sizeof(f));//初始化
for(i = 1; i <= n; i ++)
{
char ch = 'A';
for(j = 1; j <= m; j ++)
str[i][j] = ch ++;
}
l[1].x = 1;
l[1].y = 'A';
map[1][1] = 1;//将初识下标初始化为访问过
dfs(1,1,2);
printf("Scenario #%d:\n",++t2);
if(!flag)
printf("impossible\n\n");
else
{
for(i = 1; i <= n*m; i ++)
printf("%c%d",f[i].y ,f[i].x );
printf("\n\n");
}
}
return 0;
}