深搜水题poj2488

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

题意：要求从一个点出发，向中国象棋里马那样只能走“日”形对角，问是否能走遍全场，如果不行，输出“impossible”，否则按照字典序输出路径。

题解：首先审题，明显是深搜，既然深搜，那就简单了，dfs函数，每次调用给出现在所在的点以及到达的点的个数，如果已经全部到达，标记，结束递归。

注意字典序。那个字典序必须决定于移动数组move1和move2，所以一开始没a，就是字典序搞错了。代码如下

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <queue>#include <sstream>#include <fstream>#include <set>#include <map>#define INF 1e9#define M_PI 3.14159265358979323846using namespace std;//ifstream infile;int n,m,flag;int maze[30][30];int path1[30][30],path2[30][30];int move1[8] = {-1,1,-2,2,-2,2,-1,1},move2[8] = {-2,-2,-1,-1,1,1,2,2};void dfs(int x,int y,int cnt){    if(cnt == n*m){        flag = 1;        return;    }    for (int i = 0;i < 8;i++)    {        int nx = x+move1[i],ny = y+move2[i];        if((nx>=0&&nx<n)&&(ny>=0&&ny<m)&&maze[nx][ny]==0)        {            maze[nx][ny] = 1;            path1[x][y] = nx,path2[x][y] = ny;            dfs(nx,ny,cnt+1);            if(flag)            {                return;            }            path1[x][y] = -1,path2[x][y] = -1;            maze[nx][ny] = 0;        }    }}void print(int x,int y){    if(path1[x][y]==-1&&path2[x][y]==-1){        cout<<(char)(y+'A')<<x+1;        return;    }    cout<<(char)(y+'A')<<x+1;    print(path1[x][y],path2[x][y]);}int main(void){    //infile.open("1234.txt",ios::in);    int t,k = 1;    cin>>t;    while (t--)    {        flag = 0;        cin>>n>>m;        memset(path1,-1,sizeof(path1));        memset(path2,-1,sizeof(path2));        memset(maze,0,sizeof(maze));        maze[0][0] = 1;        dfs(0,0,1);        printf("Scenario #%d:\n",k++);        if(!flag){            cout<<"impossible"<<endl<<endl;            continue;        }        print(0,0);        cout<<endl<<endl;    }    return 0;}