深搜：A Knight's Journey

A Knight's Journey

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 40   Accepted Submission(s) : 13

Problem 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 <cstdio>#include <cstring>#include <algorithm>#include <string>#define Max 30using namespace std;int p,q;int visit[Max][Max];int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};int dfs(int x,int y,int step,string ans){    if(step==p*q){        cout<<ans<<endl<<endl;        return 1;    }    else{        for(int i=0;i<8;i++){            int x1=x+dir[i][0];            int y1=y+dir[i][1];            char ans1=y1+'A';            char ans2=x1+'1';            if(x1>=0&&y1>=0&&x1<p&&y1<q&&visit[x1][y1]==0){                visit[x1][y1]=1;                if(dfs(x1,y1,step+1,ans+ans1+ans2))                    return 1;                visit[x1][y1]=0;            }        }        return 0;    }}int main(){    int n,no=0;    cin>>n;    while(n--){        no++;        memset(visit,0,sizeof(visit));        visit[0][0]=1;        cin>>p>>q;        cout<<"Scenario #"<<no<<":"<<endl;        if(dfs(0,0,1,"A1")==0)            cout<<"impossible"<<endl<<endl;    }    return 0;}