1. A Knight's Journey

骑士游历问题

题目链接

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

题意分析：
基础dfs暴力，类似于马走日问题，但是要注意几个问题：
1.判断不能全部遍历完的情况 输出impossible
2.每一次输出要输出两个“\n”
3.输出第一次以字典序最小排序的遍历情况

代码： 

#include<stdio.h>
#include<stdbool.h>
#include<string.h>
//通过控制遍历方向（前进一格优先，后退两个优先）使字典序最小排序
int solution_x[8]= {-1,1,-2,2,-2,2,-1,1};
int solution_y[8]= {-2,-2,-1,-1,1,1,2,2};
int a[1000];
char b[1000];
int T,n,m,flag;
int chess[10][10]= {0};  //模拟棋盘
int count;
void SetChess(int step,int x1,int y1)
{
    int i;
    if(step==n*m)
    {
        if(count==0)  //count记录第几个解 若是第一个 则输出
        {
            printf("A1");
            for(i=1; i<step; i++)
                printf("%c%d",b[i],a[i]);
            printf("\n\n");
            flag=1;
            count++;
        }
        return;
    }
    for (i=0; i<8; i++)
    {
        int x=x1+solution_x[i],y=y1+solution_y[i];
        if (chess[x][y]==0&& x>0&&y>0 && x<=n&&y<=m)
        {
            chess[x][y]=1;
            a[step]=x,b[step]='A'+y-1; 
            SetChess(step+1,x,y);
            chess[x][y]=0;
        }
    }
}
int main()
{
    scanf("%d",&T);
    int number=1;
    while (T--)
    {
        count=0;
        flag=0;
        scanf("%d %d",&n,&m);
        printf("Scenario #%d:\n",number++);
        memset(chess,0,sizeof(chess));
        chess[1][1]=1;
        SetChess(1,1,1);
        if(flag==0)
            printf("impossible\n\n");
    }
}

 

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