poj2488 bfs

http://poj.org/problem?id=2488

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41951		Accepted: 14274

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

3

1 1

2 3

4 3

Sample Output

Scenario #1:

A1

Scenario #2:

impossible

Scenario #3:

A1B3C1A2B4C2A3B1C3A4B2C4

 

记录路径。

 

AC代码：

#include<cstdio>
#include<cstring>
using namespace std;

int T,p,q;
int To[][]= {{-,-},{-,},{-,-},{-,},{,-},{,},{,-},{,}};
int trp[][],visit[][];

int dfs(int ed,int be,int x,int y)
{
    if(be==ed)return ;
    for(int i=; i<; i++)
    {
        int xx=x+To[i][];
        int yy=y+To[i][];
        if(xx>= && xx<q && yy>= && yy<p && !visit[xx][yy])
        {
            visit[xx][yy]=;
            if(dfs(ed,be+,xx,yy))
            {
                trp[be][]=xx;
                trp[be][]=yy;
                return ;
            }
            visit[xx][yy]=;
        }
    }
    return ;
}

int main()
{
    int k=;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&p,&q);
        memset(visit,,sizeof(visit));
        trp[][]=;
        trp[][]=;
        visit[][]=;
        printf("Scenario #%d:\n",++k);
        if(dfs(p*q,,,))
        {
            for(int i=; i<p*q; i++)
                printf("%c%d",trp[i][]+'A',trp[i][]+);
        }
        else
            printf("impossible");
        printf("\n");
        if(T!=)printf("\n");
    }
    return ;
}