A Knight's Journey [dfs]

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 Input31 12 34 3Sample OutputScenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

解题报告

字典序，只要控制转移方向的顺序即可

#include<stdio.h>#include<string.h>#define MAX_N 30bool vis[MAX_N][MAX_N];int W,H,goal;int oy[]={0,-2,-2,-1,-1,1,1,2,2};int ox[]={0,-1,1,-2,2,-2,2,-1,1};char str[130];bool dfs(int X,int Y,int cnt){    if(cnt==goal) return true;    for(int i=0;i<9;i++){        int x=X+ox[i];        int y=Y+oy[i];        if(0<x&&x<=W&&0<y&&y<=H&&!vis[x][y]){            vis[x][y]=true;            if(dfs(x,y,cnt+1)){                str[2*cnt]='A'+y-1;                str[2*cnt+1]='0'+x;                return true;            }            vis[x][y]=false;            //加上这个条件32ms 立马变0ms            if(cnt==0) return false;        }    }    return false;}int main(){    int T;    scanf("%d",&T);    for(int t=1;t<=T;t++){        memset(vis,false,sizeof(vis));        scanf("%d%d",&W,&H);goal=W*H;        printf("Scenario #%d:\n",t);        if(dfs(1,1,0)){            str[W*H*2]=0;            puts(str);        }        else puts("impossible");        putchar('\n');    }    return 0;}