poj--A Knight's Journey

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本文探讨了骑士周游问题的解决方法，采用深度优先搜索（DFS）算法找到骑士访问棋盘上每个格子恰好一次的路径。文章通过具体实例展示了如何通过调整搜索方向来优化路径顺序，并提供了实现代码。

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一道经典的搜索，回溯题吧。

深搜有点想树的前序遍历。方向数组的顺序在这里非常重要。要优先左上方向。

还有今天发现以前一直理解错误的一个地方，只要是dfs搜到了目标就能输出那条路径，没有想bfs那样麻烦。

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

A Knight's Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 41240 Accepted: 14034

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

<span style="font-family:Arial;">#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int mark[30][30],n,m,B,path[1005][2],j;
<span style="color:#ff0000;">int F[8][2]= {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2},{1, -2}, {1, 2}, {2, -1}, {2, 1}};//没有做这题还真的想不到方向数组还能决定路径的顺序。注意我是y,x的方向。</span>
void dfs(int x,int y,int t)
{
    if(B)
        return;
<span style="color:#33ff33;"> </span><span style="color:#33cc00;">  </span><span style="color:#3366ff;"> <span style="background-color: rgb(255, 255, 255);">path[t][0]=x;///记住寻找的路径，不用注意位置，不然会记错
    path[t][1]=y;///</span></span>
    if(t==n*m)
    {
        B=1;
        return;
    }
   <span style="color:#ff0000;"> mark[x][y]=1;</span>
    for(int i=0; i<8; i++)
    {
        int r=F[i][1]+x;
        int c=F[i][0]+y;
        if(r>0&&r<=n&&c>0&&c<=m&&!mark[r][c])
        {
            dfs(r,c,t+1);
        }
    }
    <span style="color:#ff0000;">mark[x][y]=0;</span>
}
int main()
{
    int T,t;
    scanf("%d",&T);
    for(t=1; t<=T; t++)
    {
        memset(path,0,sizeof(path));
        j=1;
        memset(mark,0,sizeof(mark));
        scanf("%d %d",&n,&m);
        B=0;
        dfs(1,1,1);
        cout<<"Scenario #"<<t<<":"<<endl;
        if(B)
        {
            for(int i=1; i<=n*m; i++)
            {
                char s=path[i][1]+64;
                cout<<s<<path[i][0];
            }
            cout<<endl;
        }
        else
        {
            cout<<"impossible"<<endl;
        }
        if(t!=T)
            cout<<endl;
    }
    return 0;
}</span>

///--------------------------------------------------------------------------------

代码二：

<span style="font-family:Arial;">#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int mark[30][30],n,m,B,path[1005][2],j;
<span style="background-color: rgb(255, 255, 0);">int F[8][2]= {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2},{1, -2}, {1, 2}, {2, -1}, {2, 1}};//这里在尝试方向数组的顺序，yx
int F[8][2]= {{-1,-2},{1,-2},{-2, -1},{2, -1},{-2, 1},{2, 1}, {-1, 2},{1, 2}};//xy
int F[8][2]= {{1,-2},{-1,-2},{2, -1},{-2, -1},{2, 1},{-2, 1}, {1, 2},{-1, 2}};</span>
void dfs(int x,int y,int t)
{
    if(B)
        return;
    if(t==n*m)
    {
        B=1;
        return;
    }
    for(int i=0; i<8; i++)
    {
        int r=F[i][0]+x;
        int c=F[i][1]+y;
        if(r>0&&r<=n&&c>0&&c<=m&&!mark[r][c])
        {
            mark[r][c]=1;
            dfs(r,c,t+1);
            mark[r][c]=0;
            if(B)
            {
                path[t+1][0]=r;
                path[t+1][1]=c;
                return;
            }

        }
    }

}
int main()
{
    int T,t;
    scanf("%d",&T);
    for(t=1; t<=T; t++)
    {
        memset(path,0,sizeof(path));
        j=1;
        memset(mark,0,sizeof(mark));
        scanf("%d %d",&n,&m);
        B=0;
        path[1][0]=1;
        path[1][1]=1;
        mark[1][1]=1;
        dfs(1,1,1);
        cout<<"Scenario #"<<t<<":"<<endl;
        if(B)
        {
            for(int i=1; i<=n*m; i++)
            {
                char s=path[i][1]+64;
                cout<<s<<path[i][0];
            }
            cout<<endl;
        }
        else
        {
            cout<<"impossible"<<endl;
        }
        if(t!=T)
            cout<<endl;
    }
    return 0;
}</span>