【dfs】POJ-2488-A Knight's Journey

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本文探讨了一个经典的计算机科学问题——骑士周游问题。该问题要求在一个指定大小的国际象棋棋盘上找到一条路径，使得骑士能够恰好访问每个方格一次。文章提供了问题描述、输入输出格式、样例以及一种基于深度优先搜索的解决方案。

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A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 48296		Accepted: 16393

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

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

代码：

dfs+字典排序（方向！dfs过程！）

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

int dx[8]={-1,1,-2,2,-2,2,-1,1};
int dy[8]={-2,-2,-1,-1,1,1,2,2};
struct node
{
    int x;
    int y;
} a[111];

int vis[1111][1111];
int t,p,q;
int flag;

void dfs(int x,int y,int step)
{
    vis[x][y]=1;
    a[step].x=x;
    a[step].y=y+'A'-1;
    if(step==p*q)
    {
        flag=1;
        return ;
    }
    for(int i=0; i<8; i++)
    {
        int nex=x+dx[i];
            int ney=y+dy[i];
        if(nex>=1&&nex<=p&&ney>=1&&ney<=q&&vis[nex][ney]==0&&!flag)
        {
            dfs(nex,ney,step+1);//看下一个位置
            vis[nex][ney]=0;//撞完南墙回到这里，把刚用过的数字收回
        }

    }
    // vis[x][y]=0;
}




int main()
{

    scanf("%d",&t);
    for(int i=1; i<=t; i++)
    {
        flag=0;
        scanf("%d%d",&p,&q);
        memset(vis,0,sizeof(vis));
        dfs(1,1,1);
        printf("Scenario #%d:\n",i);
        if(flag)
        {
            for(int i=1; i<=p*q; i++)
            {
                printf("%c%d",a[i].y,a[i].x);
            }
            printf("\n");
        }
        else
        {
            printf("impossible\n");
        }
        if(i!=t)
            printf("\n");
    }
}