搜索练习题B-02

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

题目大意就是给出棋盘长宽，求是否能遍历棋盘，如果能要输出遍历顺序，输出按照字典序最小的；

看到题时第一反应无非是广搜，但提交数次都没过后发现字典序的问题，才发现dx与dy要按一定顺序（看代码）才能保证字典序，这样才正确，细节看代码吧。

#include<iostream>#include<string>#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>using namespace std;int vis[27][27],p,q,o=1,ok=0,dy[8]={-2,-2,-1,-1,1,1,2,2},dx[8]={-1,1,-2,2,-2,2,-1,1};//顺序必须这样从下到上，从左到右。struct node{    int x,y;//位置}an[200];void print()//打印{    char a;    for(int i=0;i<o;i++)    {        a='A'-1+an[i].y;        cout<<a<<an[i].x;    }}void solve(int x,int y){    if(ok) return ;//结束    if(o==p*q) {ok=1;print();return ;}//全部遍历完，ok=1，结束    for(int i=0;i<8;i++)    {        if(x+dx[i]>0&&x+dx[i]<=p&&y+dy[i]>0&&y+dy[i]<=q)//判断棋盘范围之内        {            if(!vis[x+dx[i]][y+dy[i]])            {                vis[x+dx[i]][y+dy[i]]=1;//标志访问过                an[o].x=x+dx[i];an[o].y=y+dy[i];o++;                solve(x+dx[i],y+dy[i]);                vis[x+dx[i]][y+dy[i]]=0;o--;//不能忘记恢复            }        }    }    return ;//如果无法遍历，也要结束，这步不能少}int main(){    int z=1,n;    cin>>n;    while(n--)    {        cin>>p>>q;        vis[1][1]=1;        an[0].x=1;an[0].y=1;        cout<<"Scenario #"<<z<<":"<<endl;z++;        solve(1,1);        if(!ok) cout<<"impossible";        cout<<endl;        if(n) cout<<endl;ok=0;o=1;    }}