poj 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 38317		Accepted: 13004

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

这道题有不少需要注意的地方；

首先注意字母表示的是列号，数字表示的是行号；

因为要按照字典序输出，注意搜索的顺序；

找到答案后，立即退出，小心覆盖答案；

#include<iostream>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<stack>
#include<queue>
#include<iomanip>
#include<map>
#include<set>
#include<functional>
#define pi 3.14159265358979323846
using namespace std;
int p,q;
int len;
struct Point
{
    int x;
    int y;
}point[100];
bool vis[30][30];
bool flag=0;
int dx[]={-1,1,-2,2,-2,2,-1,1};
int dy[]={-2,-2,-1,-1,1,1,2,2};
void dfs(int x,int y)
{
    if(len==p*q)
    {
        flag=1;
        //cout<<'A'<<endl;
        return ;
    }
    for(int i=0;i<8;++i)
    {
        int tx=x+dx[i];
        int ty=y+dy[i];
        if(tx>=0&&tx<p&&ty>=0&&ty<q&&vis[tx][ty]!=1)
        {
            vis[tx][ty]=1;
            point[len].x=tx;
            point[len].y=ty;
            ++len;
            /*for(int j=0;j<p*q;++j)
            {
                printf("%d%d ",point[j].x,point[j].y);
            }
            printf("\n");
            cout<<len<<" "<<flag;
            printf("\n");*/
            dfs(tx,ty);
            if(len==p*q)
            {
                flag=1;
                //cout<<'B'<<endl;
                return;
            }
            --len;
            vis[tx][ty]=0;
        }
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    int cnt=0;
    while(n--)
    {
        len=0;
        ++cnt;
        flag=0;
        memset(vis,0,sizeof(vis));
        scanf("%d %d",&p,&q);
        point[0].x=0;
        point[0].y=0;
        vis[0][0]=1;
        ++len;
        dfs(0,0);
        printf("Scenario #%d:\n",cnt);
        if(flag==0)
            printf("impossible");
        else
        {
            for(int i=0;i<p*q;++i)
                printf("%c%d",(point[i].y+'A'),(point[i].x+1));
        }
        printf("\n\n");
    }
    return 0;
}