POJ 2488A Knight's Journey(搜索入门)

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 46861 Accepted: 15959

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

题意：给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

想法：深搜

代码：

#include <iostream>#include<stdio.h>#include <cstring>using namespace std;const int N = 35;int vis[N][N];int p, q, flag;char b[500];int dir[8][2] = {-1, -2, 1, -2, -2, -1, 2, -1, -2, 1, 2, 1, -1, 2, 1, 2};void dfs(int x, int y, int cnt){    if (cnt == p*q)    {        printf("A1");        for (int i = 2; i < 2*cnt; i++)           printf("%c",b[i]);        printf("\n\n");        flag = 1;        return;    }    for (int i = 0; i < 8 && !flag; i++)    {       int tx = x + dir[i][0], ty = y + dir[i][1];       if (tx < 1 || tx > p || ty < 1 || ty > q)        continue;       if (!vis[tx][ty])       {           vis[tx][ty] = 1;           b[2*cnt] = 'A' + ty - 1;           b[2*cnt+1] = '1' + tx - 1;           dfs(tx, ty, cnt+1);           vis[tx][ty] = 0;        }    }}int main(){    int t, ca = 0;    scanf("%d",&t);    while (t--)    {        scanf("%d %d",&p,&q);        printf("Scenario #%d:\n",++ca);        memset(vis, 0, sizeof(vis));        flag = 0;        vis[1][1] = 1;        dfs(1, 1, 1);        if (!flag) printf("impossible\n\n");    }    return 0;}