POJ 2488 A Knight's Journey

POJ 2488 A Knight’s Journey

Description

BackgroundThe knight is getting bored of seeing the same black and white squares again and again and has decided to make a journeyaround 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.

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

这道题是说，让Knight走遍棋盘上的每一个点，但不重复。最重要的一点是，输出的时候需要按照字典序输出！

按字典序！！这该怎么输出呢？？
当然有办法输出了    (╯▽╰)
只要深搜的按照从左到右，从上到下的顺序进行搜索，并标记搜索过的地方；就可以了     o(*￣︶￣*)o
另外还有一点，国际象棋的棋盘－－横的是字母，竖的是数字 (注意:一定要按照这个格式，否则即使是答案对，也WA)
代码如下     ～(￣▽￣～)(～￣▽￣)～

 #include<iostream> #include <cstdio> #include <string.h> #include <queue>using namespace std;int n, num, alp;bool flag = false;int x[100], y[100];int vis[100][100];int dir[8][2] = {{-1, -2}, {1, -2}, {-2, -1}, {2, -1}, {-2, 1}, {2, 1}, {-1, 2}, {1, 2}};void dfs(int x1, int y1, int step){    x[step] = x1;    y[step] = y1;    if (step == num * alp)    {        flag = true;        return;    }    for(int i = 0; i < 8; i++)    {        int nx = x1 + dir[i][0];        int ny = y1 + dir[i][1];        if (nx >= 1 && nx <= num && ny >= 1 && ny <= alp && !vis[nx][ny] && !flag)        {            vis[nx][ny] = 1;            dfs(nx, ny, step + 1);            vis[nx][ny] = 0;        }    }}int main(){ #ifndef  ONLINE_JUDGE    freopen("1.txt", "r", stdin); #endif    cin >> n;    for (int i = 1; i <= n; i++)    {        cin >> num >> alp;        flag = false;        memset(x, 0, sizeof(x));        memset(y, 0, sizeof(y));        memset(vis, 0, sizeof(vis));        vis[1][1] = 1;        dfs(1, 1, 1);        printf("Scenario #%d:\n",i);        if (flag)        {            for (int i = 1; i <= num * alp; i++)            {                printf("%c%d", y[i]+'A' - 1, x[i]);            }            cout << endl;        }        else        {            cout << "impossible" << endl;        }        if (i != n)        {            cout << endl;        }    }}