POJ 2488A Knight's Journey9（DFS）

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A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42414 Accepted: 14423

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：王子的在p*q的矩阵里的运动轨迹，王子的运动方式类似于象棋里的“马走日”。并且按照优先字典序的方式进行移动。

题解：简单的dfs，定义一个Map二维数组用来储存王子可以移动的八种方式，xx和yy数组用来储存每一步移动的地址（这是关键的一步）。

注意：ABC...是列，123...是行，所以注意输入。

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;int p, q, num, flag ;char Map[8][2] = {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2}, {1, -2}, {1, 2}, {2, -1}, {2, 1}};int xx[30], yy[30];int div[30][30];void dfs(int x, int y, int num){  int i, j, xi, yi;  xx[num] = x;  yy[num] = y;  if(num == p * q)    {      flag = 1;      return ;    }  for(i = 0; i < 8; i++)    {      xi = x + Map[i][0];      yi = y + Map[i][1];      if(xi >= 1 && xi  <= p && yi >= 1 && yi <= q&&div[xi][yi] == 0 && flag == 0)        {              div[xi][yi] = 1;              dfs(xi, yi, num + 1);              div[xi][yi] = 0;        }    }}int main(){  int i, j;  int n;  int cesa = 0;  cin>>n;  while(n--)    {      cesa++;      cin>>q>> p;      flag = 0;      memset(div, 0 ,sizeof(div));      div[1][1] = 1;      dfs(1,1,1);      cout<<"Scenario #"<<cesa<<":"<<endl;      if(flag == 1)        {          for(i = 1; i <= p * q; i++)            {              printf("%c%d",xx[i]+'A'-1,yy[i]);            }          cout<<endl<<endl;        }      else        {          cout<<"impossible"<<endl<<endl;        }    }  return 0;}