poj 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20870		Accepted: 7050

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

也是一道经典的搜索题，只要注意是按照字典序输出。

#include<iostream>
#include<string.h>
#include<stdio.h>
#define N 27

using namespace std;

int path
[2];

int flag,a,b;
int map

;

int move[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};//按照字典序排列

void dfs(int i,int j,int step)
{
if(step==a*b)//成功条件
{
for(int i=0;i<step;i++)
printf("%c%d",path[i][0]+'A',path[i][1]+1);
printf("\n");
flag=1;
}
else
for(int x=0;x<8;x++)
{
int n=i+move[x][0];
int m=j+move[x][1];
if(n<b&&m<a&&n>=0&&m>=0&&!map
[m]&&!flag)//在边界内且没有被访问
{
map
[m]=1;
path[step][0]=n,path[step][1]=m;
//  cout<<path[step][0]<<path[step][1]<<endl;
dfs(n,m,step+1);
map
[m]=0;//回溯
}
}
}

int main()
{
int k=1,n;
scanf("%d",&n);
while(n--)
{
flag=0;
scanf("%d %d",&a,&b);
for(int i=0;i<a;i++)
for(int j=0;j<b;j++)
map[i][j]=0;//初始化地图
map[0][0]=1;
path[0][0]=0,path[0][1]=0;//初始化起点
cout<<"Scenario #"<<k++<<":"<<endl;
dfs(0,0,1);//搜索
if(!flag)
cout<<"impossible"<<endl;
cout<<endl;
}
return 0;
}