poj 2488 A Knight's Journey

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

poj 搜索题，骑士游历问题稍稍加强版，加了地图变换和字典序的问题，深搜就可以解决，而且不需要剪枝。附上代码

#include<iostream>#include<vector>#include<stdio.h>#include<cstring>using namespace std;struct lnode{int x,y;};lnode node[10001];int total,mapx,mapy;int map[101][101];vector<lnode>v;int xx[8]={-1,1,-2,2,-2,2,-1,1};int yy[8]={-2,-2,-1,-1,1,1,2,2};bool dfs(int x,int y,int count){if(count==mapx*mapy)return true;for(int i=0;i<8;i++){int nowx,nowy;nowx=x+xx[i];nowy=y+yy[i];lnode ans;ans.x=nowx;ans.y=nowy;if(nowx>0&&nowx<=mapx&&nowy>0&&nowy<=mapy&&map[nowx][nowy]==0){v.push_back(ans);map[nowx][nowy]=1;if(dfs(nowx,nowy,count+1))return true;map[nowx][nowy]=0;v.pop_back();

//这个地方加了一个return false;错了N次}}return false;}int main(){int t;cin>>t;for(int i=1;i<=t;i++){v.clear();memset(map,0,sizeof(map));cin>>node[i].x>>node[i].y;total=node[i].x*node[i].y;mapx=node[i].x;mapy=node[i].y;map[1][1]=1;lnode L;L.x=L.y=1;v.push_back(L);printf("Scenario #%d:\n",i);if(dfs(1,1,1)){for(int i=0;i<v.size();i++)cout<<(char)(v[i].y+'A'-1)<<v[i].x;cout<<endl;}else cout<<"impossible"<<endl;cout<<endl;}return 0;}