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A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 25493 Accepted: 8686

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

这是一道深搜的题目，主要是要回溯，这个题还有就是要按字典序来输出，所以方向数组就不能像以前那样写了，要按8个方向的字典序搜索，我表示我被坑的是格式居然错了，老是pe，最后才发现少了一个：，以后一定要注意细节。

下面是代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cstring>#include <vector>#include <cassert>#include <cstdlib>using namespace std;const int maxn=1000;int map[maxn][maxn];int sx[maxn],sy[maxn];int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};//这点一定要注意，我写的时候不仔细老是waint p,q,flag,step;void init(){    for(int i=1;i<=p;i++)        for(int j=1;j<=q;j++)             map[i][j]=0;}void dfs(int i,int j){     if(flag)       return;     int x,y,k;     step++;     sx[step]=i;     sy[step]=j;     if(step==p*q)     {          flag=1;          return;     }     map[i][j]=1;     for(int k=0;k<8;k++){          y=j+dir[k][0];          x=i+dir[k][1];          if(map[x][y]==0&&x>0&&x<=p&&y>0&&y<=q){                dfs(x,y);                step--;          }     }     map[i][j]=0;}int main(){    int n,count=0;    cin>>n;    while(n--){        step=0;        flag=0;        count++;        cin>>p>>q;        init();        dfs(1,1);        cout<<"Scenario #"<<count<<":"<<endl;        if(flag){            for(int i=1;i<=p*q;i++)               printf("%c%d",sy[i]+64,sx[i]);//表示这一点输出是从网上的大牛那里学到的               printf("\n");        }        else            cout<<"impossible"<<endl;        if(n!=0) cout<<endl;//这点要注意最后一行是没空行的，不然会pe    }    return 0;}