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[N][N],m,n,flag;
char str[N][N];
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:
",++t2);
        if(!flag)
            printf("impossible

");
        else
        {
            for(i = 1; i <= n*m; i ++)
                printf("%c%d",f[i].y ,f[i].x );
            printf("

");
        }
    }
    return 0;
}