POJ 2488 A Knight's Journey (dfs)

POJ 2488 A Knight's Journey (dfs)

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

马要走遍p*q的棋盘的所有格，它可以从任意点出发，问能否成功，若能，输出所能走的最小字典序的路径。

思路

简单dfs，判断马是否走遍了所有格我们可以用搜索的深度来判断，如果它等于棋盘格的数量，则已经走遍了所有，在dfs过程中，我们要保证走过的点不会被重复踏入，另外一个便是如何保证字典序最小，可以采用搜索的顺续来确定，具体顺续见move数组。

AC 代码

#include #include #include #include #include #include#include using namespace std;bool visited[30][30],flag;int move[8][2]= {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};int p,q;struct node{ char x; char y; void init(int x,int y) { this->x=x; this->y=y; }} sk[1005];void dfs(int x,int y,int deep){ if(deep==p*q) //已经走遍了所有格 { for(int i=0; i=p||yi<0||yi>=q)continue; //棋盘外 if(visited[xi][yi]==false) //该位置为空 { visited[xi][yi]=true; sk[deep].init(yi+'A',xi+'1'); dfs(xi,yi,deep+1); if(flag)return; visited[xi][yi]=false; } }}int main(){ int n; scanf("%d",&n); for(int ni=1; ni<=n; ni++) { printf("Scenario #%d:\n",ni); scanf("%d%d",&p,&q); flag=false; //当前未找到路径 memset(visited,false,sizeof(visited)); for(int j=0; j

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