ACM-搜索-02

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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

题目思路：

马的遍历问题，使用dfs，可以跳跃八个方向

x[8]={-1, 1, -2, 2, -2, 2, -1, 1},y[8]={-2, -2, -1, -1, 1, 1, 2, 2};

这个方向的顺序是固定的，因为题目中说输出的这一条路径是以字典为基础的，其实就是跳越的顺序是确定的，第一次没有注意到这一点，以为随便输出一条j就ok了，记过显示错误答案。

每次跳跃时要进行判断，是否到边界，是否要跳到的点已经使用过，如果能跳，则改变坐标，并标记已经走过。当移动步数等于方格数p*q时说明已经遍历了，则输出；

代码如下：

#include<string.h>#include<cstdio>#include<iostream>#include<cstdlib>using namespace std;int a[100][2],b[9][9],t=0,p,q,flag,n;int x[8]={-1, 1, -2, 2, -2, 2, -1, 1},y[8]={-2, -2, -1, -1, 1, 1, 2, 2};char A='A';void search(int ax,int ay,int i){    a[i][0]=ax;    a[i][1]=ay;    if(i==p*q){flag=1;return;}    for(int j=0;j<8;j++)    if(ax+x[j]>=1&&ax+x[j]<=p&&ay+y[j]>=1&&ay+y[j]<=q&&b[ax+x[j]][ay+y[j]]!=0&&flag==0)    {        int newax=ax+x[j];        int neway=ay+y[j];        b[newax][neway]=0;        search(newax,neway,i+1);        b[newax][neway]=1;    }}int main(){    cin>>n;    while(n--)    {        cin>>p>>q;        t++;flag=0;        memset(b,1,sizeof(b));        memset(a,0,sizeof(a));        b[1][1]=0;        search(1,1,1);        cout<<"Scenario #"<<t<<":"<<endl;        if(flag)        {                for(int j=1;j<=p*q;j++)                {                    A='A';                    A=A+-1+a[j][1];                    cout<<A<<a[j][0];                    }                cout<<endl;            }        else            cout<<"impossible"<<endl;            cout<<endl;        }}