POJ 2488-A Knight's Journey（DFS）

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 31702 Accepted: 10813

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：国际象棋，然后给一个马（马走日） ，可以从任意点出发，找一条可以访问所有格子（p*q的棋盘）的路径，注意路径如果有多条要求输出字典序最小的那条。。然后这个可以搜索的时候按字典序搜。。就是搜索方向要固定。。不能随意写了

然后其他的没什么了 直接深搜，搜到答案之后直接return ；

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cctype>#include <vector>#include <cstdio>#include <cmath>#include <deque>#include <stack>#include <map>#include <set>#define ll long long#define maxn 116#define pp pair<int,int>#define INF 0x3f3f3f3f#define max(x,y) ( ((x) > (y)) ? (x) : (y) )#define min(x,y) ( ((x) > (y)) ? (y) : (x) )using namespace std;int n,m,k,ans,dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};bool vis[27][27];int sx[30],sy[30],top,ok;void dfs(int x,int y){if(ok) return ;if(top==n*m){ok=1;for(int i=0;i<top;i++)printf("%c%d",'A'+sy[i]-1,sx[i]);return ;}for(int i=0;i<8;i++){int tx=x+dir[i][0];int ty=y+dir[i][1];if(tx>=1&&tx<=n&&ty>=1&&ty<=m&&!vis[tx][ty]){vis[tx][ty]=1;sx[top]=tx;sy[top++]=ty;dfs(tx,ty);vis[tx][ty]=0;top--;}}}int main(){int T,cas=1;scanf("%d",&T);    while(T--){scanf("%d%d",&n,&m);ok=0;printf("Scenario #%d:\n",cas++);memset(vis,0,sizeof(vis));top=0;vis[1][1]=1;sx[top]=1;sy[top++]=1;dfs(1,1);if(!ok)printf("impossible");puts("");if(T)puts("");}return 0;}