POJ 2488 A Knight's Journey (DFS + 记录路径)

A Knight's Journey

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 32388
Accepted: 11027

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

题目链接：http://poj.org/problem?id=2488

题目大意：给一个p*q的棋盘，行用数字表示，列用字母表示，求用马步遍历的遍历序列，马走“日”。

题目分析：没啥好说的，裸DFS，因为题目说了从任意一点开始都可以，说明解是一个圈，所以当然从A1开始搜字典序最小了，方向遍历时也要根据字典序来遍历，关键是记录路径，和k题类似，二维数组，一维用深度来记录路径

#include <cstdio>#include <cstring>int vis[30][30];int path[30][2];int p,q;bool flag;int dirx[8] = {-1,1,-2,2,-2,2,-1,1};int diry[8] = {-2,-2,-1,-1,1,1,2,2};void DFS(int x,int y, int step){    path[step][0] = x;    path[step][1] = y;    if(step == p * q)    {        flag = true;        return;    }    for(int i = 0; i < 8; i++)    {        int xx = x + dirx[i];        int yy = y + diry[i];        if(xx < 1 || yy < 1 || xx > p || yy > q || vis[xx][yy] || flag)            continue;           vis[xx][yy] = 1;        DFS(xx, yy, step + 1);        vis[xx][yy] = 0;    }}int main(){    int T;    scanf("%d",&T);    for(int i = 1; i <= T; i++)    {        scanf("%d %d",&p, &q);        memset(vis, 0, sizeof(vis));        vis[1][1] = 1;        flag = false;        DFS(1, 1, 1);        if(flag)        {            printf("Scenario #%d:\n",i);            for(int i = 1; i <= p * q; i++)                printf("%c%d", path[i][1] - 1 + 'A', path[i][0]);            printf("\n");        }        else            printf("Scenario #%d:\nimpossible\n",i);        if(i != T)            printf("\n");    }}