POJ 2488 A Knight's Journey（dfs）

题目代号：POJ 2488

题目链接：http://poj.org/problem?id=2488

 

 

Language: Default

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46017		Accepted: 15657

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

题目大意：国际象棋中的马只能走的路径如图所示，第一列的位置是A，第一行的位置是1，所以1,1的点位置是A1，以此类推，现在让你求能否有一种方式能走完棋盘上所有的点，如果有则输出路径中按字典序排最小的那一种路径。

题目思路：假设一个位置有8种方向可以选择，那么把8个方向对应的点的坐标按字典序排就好了，顺序一定不能错，然后dfs遍历一遍就行了，这条路径一定是字典序最小的那个。

AC代码：

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL;

int a[30][30],n,m,flag;
int cx[]={-2,-2,-1,-1,1,1,2,2};
int cy[]={-1,1,-2,2,-2,2,-1,1};
map<int,string>M;

void dfs(int i,int j,int step)
{
    if(flag)return;
    //if(i<1||i>m||j<1||j>n||a[i][j])return;
    a[i][j]=1;
    M[step]="";
    M[step]+='A'-1+i,M[step]+=j+'0';
    if(step==n*m)
    {
        flag=1;
        return;
    }
    for(int k=0;k<8;k++)
    {
        int x=i+cx[k];
        int y=j+cy[k];
        if(x>0&&x<=n&&y>0&&y<=m&&!a[x][y]&&!flag)
        {
            dfs(x,y,step+1);
            a[x][y]=0;
        }
    }
}

int main()
{
    int t,i=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        if(i)puts("");
        printf("Scenario #%d:\n",++i);
        flag=0;
        mem(a,0);
        M.clear();
        dfs(1,1,1);
        if(flag)
        {
            for(int i=1;i<=n*m;i++)
                cout<<M[i];
            cout<<endl;
        }
        else
        {
            printf("impossible\n");
        }
    }
    return 0;
}