[ACM]A Knight's Journey DFS应用

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 3Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

        很囧的一道题目,其实就是判断在制定的棋盘大小内是否会出现汉密尔顿回路,由于棋盘最大也才26格,DFS就行了,找到第一个解救输出答案,一直找不到就输出不可能.网上有的做法是从每个点都试一遍,其实完全没有必要,因为只要有的话就需要遍历每个点,此时每个点作为起点都能成功.唯一需要注意的是要记录路径,并且一个是'A'开头，一个是1开头，有点相应转换，最关键的还是需要注意尝试路径的顺序，不对的话会WA的....我反正试了几个顺序才对，莫非是题目理解错了？

#include "iostream" 
using namespace std;
//1,1肯定在回路中,所以只需要判断从1,1出发即可 
int p,q;
int str[60];
bool find1;
bool a[30][30];
const int xShift[8]={-2,-2,-1,-1,1,1,2,2};//顺序很重要....没有special judge 
const int yShift[8]={-1,1,-2,2,-2,2,-1,1};
void Dfs(const int x,const int y,const int n){
    //cout<<x<<" "<<y<<"...."<<n<<"......."<<endl; 
    if(n==p*q-1){
        //puts("if"); 
        for(int i=0;i<2*p*q-2;i+=2){
            printf("%c%d",'A'+str[i]-1,str[i+1]);
        }
        printf("%c%d/n",'A'+x-1,y);
        find1=true;
    }else if(!find1){
        //puts("else"); 
        str[n*2]=x;str[n*2+1]=y;//record; 
        a[x][y]=false;
        for(int i=0;i<8;i++){
            int X=x+xShift[i],Y=y+yShift[i];
            //cout<<X<<" .............. "<<Y<<endl; 
            if(0<X && X<=p && 0<Y && Y<=q && a[X][Y]){
                Dfs(X,Y,n+1);
                //cout<<X<<" .. "<<Y<<endl; 
            }
        }
        a[x][y]=true;
    }
}
int main(){
    int t,i,j,k;
    scanf("%d",&t);
    for(i=0;i<t;i++){
        scanf("%d%d",&q,&p);
        printf("Scenario #%d:/n",i+1);
        find1=false;
        memset(a,true,sizeof(a));
        str[0]=str[1]=1;
        Dfs(1,1,0);
        if(!find1){
            printf("impossible/n",i+1);
        }
        puts("");
    }
    return 0;
}