A Knight's Journey POJ 2488

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

#include<iostream>#include<stdio.h>#include<stack>#include<string.h>using namespace std;#define MAXN 28struct Node{    int x;    int y;};int chess[MAXN][MAXN];int p,q;Node path[MAXN*MAXN];bool flag;int dx[]={-2,-2,-1,-1,1,1,2,2};int dy[]={-1,1,-2,2,-2,2,-1,1};void solve(int x,int y,int step){    if(flag) return;    path[step].x=x;    path[step].y=y;    if(step==p*q){        flag=true;  //end of flag        return;    }    int curX,curY;    for(int a=0;a<8;a++){        curX=x+dx[a];        curY=y+dy[a];        if(curX>=1&&curX<=q&&curY>=1&&curY<=p){            if(!chess[curX][curY]){ //this step is blank                chess[curX][curY]=step+1;                solve(curX,curY,step+1);                chess[curX][curY]=0;            }        }    }}int main(){    int ncase;    cin>>ncase;    for(int i=1;i<=ncase;i++){        flag=false;        memset(chess,0,sizeof(chess));        cin>>p>>q;        chess[1][1]=1;        solve(1,1,1);        cout<<"Scenario #"<<i<<":"<<endl;        if(flag){            for(int j=1;j<=p*q;j++){ //Scenario #1:  A1B3C1A2B4C2A3B1C3A4B2C4                cout<<char('A'+path[j].x-1)<<path[j].y;            }        }else{            cout<<"impossible";        }        cout<<endl<<endl;    }    return 0;}

note1:简单的dfs，注意题干中要求输出结果按字典序排列。

这类题解题模式：

dfs(){

1.是否已经有可行值

2.记录当前值（修改公共变量）

3.当前值是否为需要的结果

4.深度遍历其他结果（①修改公共变量 ②将公共变量值还原 模拟遍历过程）

}