poj 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20955 Accepted: 7087

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

哎呀呀。。。英语不好，题目理解错误。。。把字典序的坐标当成了行坐标。。。然后dfs边界里面定义就错了。。苦WA了n次。。。虽然说是dfs入门ing，但这状况不行啊，还是要提高效率的说。

#include <iostream>#include <stdio.h>using namespace std;int map[27][27];int path[27][2];int dir[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};int a,b,flag,n,k;void dfs(int i,int j,int step){    if(step==a*b)    {        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+dir[x][0];            int m=j+dir[x][1];            if(m>=0&&n>=0&&n<b&&m<a&&!flag&&!map[n][m])            {                map[n][m]=1;                path[step][0]=n;                path[step][1]=m;                dfs(n,m,step+1);                map[n][m]=0;            }        }    }}int main(){    k=1;    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;        map[0][0]=1;        path[0][1]=0;        path[0][0]=0;        cout<<"Scenario #"<<k++<<":"<<endl;        dfs(0,0,1);        if(!flag)        cout<<"impossible"<<endl;        cout<<endl;    }    return 0;}