Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 23108 Accepted: 7819

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

读懂了题目这题就会做了，以前的马在跳的时候都是在线上，这个题是这方格内还真有些不适应，转化成在线上就好

[cpp] ?

#include ?

#include ?

int status[100][100],s; ?

int vex[]={-2,-2,2,2,-1,-1,1,1}; ?

int vey[]={-1,1,-1,1,-2,2,-2,2}; ?

char path1[1000],prepath1[1000]; ?

int path2[1000],prepath2[1000],top,n,m,key,k; ?

int main() ?

{ ?

? ? void dfs(int x,int y); ?

? ? int i,j,t,tem; ?

? ? scanf("%d",&t); ?

? ? tem=1; ?

? ? while(t--) ?

? ? { ?

? ? ? ? scanf("%d %d",&n,&m); ?

? ? ? ? memset(status,0,sizeof(status)); ?

? ? ? ? if(n==1&&m==1) ?

? ? ? ? { ?

? ? ? ? ? ? printf("Scenario #%d:\n",tem); tem++; ?

? ? ? ? ? ? printf("A1\n"); ?

? ? ? ? ? ? printf("\n"); ?

? ? ? ? ? ? continue; ?

? ? ? ? } ?

? ? ? ? for(i=1;i<=n;i++) ?

? ? ? ? { ?

? ? ? ? ? ? for(j=1;j<=m;j++) ?

? ? ? ? ? ? { ?

? ? ? ? ? ? ? ? top=0;s=1; ?

? ? ? ? ? ? ? ? memset(status,0,sizeof(status)); ?

? ? ? ? ? ? ? ? prepath1[top]='A'+j-1; ?

? ? ? ? ? ? ? ? prepath2[top++]=i; ?

? ? ? ? ? ? ? ? status[i][j]=1; ?

? ? ? ? ? ? ? ? key=0; k=0; ?

? ? ? ? ? ? ? ? dfs(i,j); ?

? ? ? ? ? ? ? ? if(k==1) ?

? ? ? ? ? ? ? ? { ?

? ? ? ? ? ? ? ? ? ? break; ?

? ? ? ? ? ? ? ? } ?

? ? ? ? ? ? } ?

? ? ? ? ? ? if(j!=m+1) ?

? ? ? ? ? ? { ?

? ? ? ? ? ? ? ? break; ?

? ? ? ? ? ? } ?

? ? ? ? } ?

? ? ? ? printf("Scenario #%d:\n",tem); tem++; ?

? ? ? ? if(i!=n+1) ?

? ? ? ? { ?

? ? ? ? ? ? for(i=0;i<=n*m-1;i++) ?

? ? ? ? ? ? { ?

? ? ? ? ? ? ? ? printf("%c%d",path1[i],path2[i]); ?

? ? ? ? ? ? } ?

? ? ? ? ? ? printf("\n"); ?

? ? ? ? }else ?

? ? ? ? { ?

? ? ? ? ? ? printf("impossible\n"); ?

? ? ? ? } ?

? ? ? ? printf("\n"); ?

? ? } ?

? ? return 0; ?

} ?

void dfs(int x,int y) ?

{ ?

? ? int i,j,xend,yend,change; ?

? ? for(i=0;i<=7;i++) ?

? ? { ?

? ? ? ? xend=x+vex[i]; yend=y+vey[i]; ?

? ? ? ? if(xend>=1&&xend<=n&?d>=1&?d<=m&&!status[xend][yend]) ?

? ? ? ? { ?

? ? ? ? ? ? s++; ?

? ? ? ? ? ? prepath1[top]=yend+'A'-1; ?

? ? ? ? ? ? prepath2[top++]=xend; ?