poj 2488 A Knight's Journey（dfs+字典序路径输出）

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题目链接：http://poj.org/problem?id=2488

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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

大致题意:

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。

代码如下：

#include <cstdio>#include <cstring>#define M 26struct node{int x, y;}w[M*M];bool vis[M][M];int p, q;int flag = 0;int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};//按此顺序搜索出来的结果就是字典序bool judge(int x, int y){if(x>=0&&x<q&&y>=0&&y<p&&!vis[x][y])return true;return false;}void dfs(int x, int y, int step){w[step].x = x,w[step].y = y;vis[x][y] = true;if(step == p*q-1){flag = 1;return ;}for(int i = 0; i < 8; i++){int dx = w[step].x+dir[i][0];int dy = w[step].y+dir[i][1];if(judge(dx,dy)){vis[dx][dy] = true;dfs(dx,dy,step+1);if(flag)//一但找到就退出搜索return;vis[dx][dy] = false;}}return ;}void print(){for(int i = 0; i < p*q; i++)//列为字母，行为数字{printf("%c%d",w[i].x+'A',w[i].y+1);}printf("\n\n");}int main(){int t, i, j, cas = 0;scanf("%d",&t);while(t--){memset(vis,false,sizeof(vis));flag = 0;scanf("%d%d",&p,&q);for(i = 0; i < q; i++)//列{for(j = 0; j < p; j++)//行{dfs(i,j,0);if(flag)break;}if(flag)break;}printf("Scenario #%d:\n",++cas);if(flag)print();elseprintf("impossible\n\n");}return 0;}