POJ 2488 A Knight's Journey (DFS深搜)

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 45901 Accepted: 15624

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

123

题目大意: 走日字形 能否走完这个圆 可以  字典序输出 

日字形 走法:

注意 方向的问题

#include <iostream>#include <queue>#include <string>#include <cstring>#include <stdio.h>#include <cmath>using namespace std;/*日字形  从(1,1) 开始, 并且 行为数字, 列为字母, 字典序 先记录列 后行*/const int N =100;int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};// 注意顺序int vis[N][N];int maps[N][N];int p,q,ans;int flag;bool judge(int x,int y){    if(x>=1&&x<=p&&y>=1&&y<=q&&!vis[x][y]&&!flag)        return true;    return false;}void dfs(int x,int y,int step){    maps[step][0]=x;    maps[step][1]=y;    if(step==p*q)    {        flag =1;        return;    }    for(int i=0;i<8;i++)    {        int px=x+dir[i][0];        int py=y+dir[i][1];        if(judge(px,py))        {            vis[px][py]=1;            dfs(px,py,step+1);            vis[px][py]=0;        }    }}int main(){    int T,cont=0;    cin>>T;    while(T--)    {        flag=0;        cin>>p>>q;        memset(vis,0,sizeof(vis));        vis[1][1]=1;        dfs(1,1,1);        printf("Scenario #%d:\n",++cont);        if(!flag)            printf("impossible");        else        {            for(int i = 1; i <= p * q; i++)                printf("%c%d",maps[i][1] - 1 + 'A',maps[i][0]);// 先列后行 列 字母  行数字        }        printf("\n");        if(T!=0)            printf("\n");    }    return 0;}