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

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

要去输出字典序最小

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N = 33;
struct P
{
    int x,y;
    P(int x,int y):x(x),y(y){}
    P(){}
}b[N*N];
char a[N][N];
int vis[N][N],n,m,flag;
int dx[]={-1,1,-2,2,-2,2,-1,1};
int dy[]={-2,-2,-1,-1,1,1,2,2};
bool ok(int x,int y)
{
    if(x<0||y<0||x>=n||y>=m) return true;
    return false;
}
void dfs(int x,int y,int cnt)
{
    int i,j,next_x,next_y;
    if(flag) return;
    if(cnt==n*m) {
        flag=1;
        for(i=1;i<=cnt;i++) printf("%c%d",b[i].y+'A',b[i].x+1);
        printf("\n");
        return ;
    }
    for(i=0;i<8;i++) {
        next_x = x + dx[i];
        next_y = y + dy[i];
        if(ok(next_x,next_y)) continue;
        if(vis[next_x][next_y]) continue;
        vis[next_x][next_y]=1;
        b[cnt+1]=P(next_x,next_y);
        dfs(next_x,next_y,cnt+1);
        if(flag) return;
        vis[next_x][next_y]=0;
    }
    return ;
}
int main()
{
    int i,j,T;
    scanf("%d",&T);
    for(int _=1;_<=T;_++) {
        if(_!=1) printf("\n");
        scanf("%d%d",&n,&m);
        memset(vis,0,sizeof(vis));
        flag=0;
        printf("Scenario #%d:\n",_);
        for(i=0;i<n;i++) {
            for(j=0;j<m;j++) {
                vis[i][j]=1;
                b[1]=P(i,j);
                dfs(i,j,1);
                vis[i][j]=0;
                if(flag) break;
            }
            if(flag) break;
        }
        if(!flag) printf("impossible\n");
    }
    return 0;
}