POJ2488——DFS——A Knight's Journey

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

大意：一个骑士从原点出发，只能走马字型路线， 要求遍历所有路径的字典序排序（就是先A再B再C），用DFS思想遍历每一种状态，如果最后步数等于图的大小那么就直接输出，难在字典序排序（虽然还不知道为什么这样写dir这个数组- -||），用visit记录已经经过的点，最后如果DFS失败那么visit要重新变成0

#include<cstdio>

#include<cstring>

using namespace std;

const int inf = 27;

int visit[inf][inf];

int dir[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};

int flag ,p, q,ans;

struct edge{

    char s;

    int num;

}a[inf*inf];

void dfs(int x,int y,int step)

{

    if(flag == 0 ) return  ;

    if(visit[x][y]) return ;

    visit[x][y] = 1;

    if(step == ans){

            flag = 0;

            for(int i = 1; i <= ans;i++)

                printf("%c%d",a[i].s,a[i].num);

                printf("\n");

                return ;



    }

    for(int i = 0; i < 8 ; i++){

        int x1 = x + dir[i][0];

        int y1 = y +dir[i][1];

        if( x1 < 1||y1 < 1||x1 >p||y1 > q||visit[x1][y1] == 1)

            continue;

        a[step+1].s = 'A' + y1 - 1;

        a[step+1].num =  x1;

        dfs(x1,y1,step+1);

        visit[x1][y1] = 0;

    }

}







int main()

{

int T,count = 1;

scanf("%d",&T);

while(T--){

    memset(visit,0,sizeof(visit));

     flag = 1;

     a[1].s = 'A',a[1].num = 1;

    scanf("%d%d",&p,&q);

    ans = p*q;

    printf("Scenario #%d:\n",count++);

    dfs(1,1,1);

    if(flag == 1)

    printf("impossible\n");

    if(T!=0)

    printf("\n");

}

return 0;

}

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