深搜2 poj 2448

poj  2448

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 21891 Accepted: 7393

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

Accepted692K16MSG++1245B输出一条遍历所有点的路径，要求每个点只能遍历一次，深搜找路径

按照字典序顺序输出遍历的路径，，dir为马走日的八个方向，按照字典序顺序遍历的八个方向为

int dx[8]={-2,-2,-1,-1,1,1,2,2},   //按字典序方向行走，用way记录走的路径，visit标记是否走过。。。
    dy[8]={-1,1,-2,2,-2,2,-1,1};

//rear是当前走的步数，下面是我徒手画的8个方向的排序

代码：

#include <cstdio>#include<iostream>#include<cstring>#include<algorithm>#define maxn 15using namespace std;int way[maxn*maxn][2],visit[maxn][maxn];int dx[8]={-2,-2,-1,-1,1,1,2,2},   //按字典序方向行走    dy[8]={-1,1,-2,2,-2,2,-1,1}; int q,p;bool dfs(int tot,int rear,int x,int y){    int i,j,nx,ny;    if(rear==tot)return 1; //搜索成功    for(i=0;i<8;i++)    {      nx=x+dir[i][0];      ny=y+dir[i][1];      if(nx>=0&&nx<q&&ny>=0&&ny<p&&!visit[nx][ny])      {          visit[nx][ny]=1; //标记为经过          if(dfs(tot,rear+1,nx,ny)) //搜索下一个          {              way[rear][0]=nx;              way[rear][1]=ny;              return 1;//搜索成功直接跳出，即为所走的路径          }          visit[nx][ny]=0; //搜索不成功，标记为未经过      }    }    return 0;//搜索不成功}int main(){    int i,j,cas=1,t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&p,&q);        memset(visit,0,sizeof(visit));        way[0][0]=0;way[0][1]=0;        visit[0][0]=1; //标记A1已经经过        printf("Scenario #%d:\n",cas++);        if(dfs(p*q,1,0,0))        {            for(i=0;i<p*q;i++)            printf("%c%d",way[i][0]+'A',way[i][1]+1);        }        else        printf("impossible");        printf("\n");        if(t!=0)printf("\n");    }}