Poj 2488 A Knight's Journey(Dfs)

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 32782 Accepted: 11154

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

题意：

给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

经典的“骑士游历”问题，DFS水题一道

#include <iostream>#include <cstdio>#include <cstring>using namespace std;struct node{    int x;    int y;}Map[26*26+1];int vis[27][27];int flag;int n,m;int dx[]={-2,-2,-1,-1,1,1,2,2};int dy[]={-1,1,-2,2,-2,2,-1,1};int ok(int x,int y){    if(x>=0&&y>=0&&x<m&&y<n&&!vis[x][y]){        return 1;    }    return 0;}void dfs(int x,int y,int step){    Map[step].x=x;    Map[step].y=y;    if(step==m*n-1){        flag=1;        return ;    }    for(int i=0;i<8;i++){        if( ok( x+dx[i],y+dy[i] ) ){            vis[ x+dx[i] ][ y+dy[i] ]=1;            dfs(x+dx[i],y+dy[i],step+1);            if(flag)                return ;            vis[ x+dx[i] ][ y+dy[i] ]=0;        }    }}int main(){    int T;    scanf("%d",&T);    for(int kk=1;kk<=T;kk++){        printf("Scenario #%d:\n",kk);        scanf("%d%d",&n,&m);        memset(vis,0,sizeof(vis));        for(int i=0;i<m;i++){            for(int j=0;j<n;j++){                vis[i][j]=1;                flag=0;                dfs(i,j,0);                if(flag)                    break;            }            if(flag)                break;        }        if(!flag){            printf("impossible\n\n");        }        else{            for(int i=0;i<m*n;i++){                printf("%c%d",Map[i].x+'A',Map[i].y+1);            }            cout<<endl<<endl;        }    }    return 0;}