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

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal 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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #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 100int 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:\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;}