A Knight's Journey (深度搜索+字典序)

A Knight's Journey

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 125   Accepted Submission(s) : 26

Problem 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.<br>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

 

Source

PKU

 

思路：

棋盘的遍历问题首先想到的是广搜，但深搜其实也适合！此题运用了一些技巧。首先单独创建一个输出函数，用来进行数据的输出，还有就是字符的处理上，将列仍然看成数字，然后在搜索函数里在转换成字符，以便存储位置。搜索的形式是（int x,int y,int step），其中xy表示几行几列（据y由y+'A'-1算出字符列），起点是(1,1,1).从头开始搜索，结束是如果x*y==step时，则搜索到结果，此时可以输出，在这个判断输出的问题上(即是否可以搜索到！)可以用一个flag标志，还有就是字典序的控制，一定要严格按照顺序进行搜索，还有就是要注意输出的格式，每组测试数据中间要有空格！

代码：

#include <stdio.h>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int dir[8][2]={-1,-2,1,-2,-2,-1,2,-1,-2,1,2,1,-1,2,1,2};   //严格按照顺序bool vis[30][30];    //判断是否走过int cas,r,c,flag=0,t;struct point{    int s;    char s1;}p[666];void print()   //输出函数{    cout<<"Scenario #"<<cas<<":"<<endl;    for(int i=1;i<=r*c;i++)    cout<<p[i].s1<<p[i].s;    cout<<endl;    if(t!=0)        cout<<endl;}bool judge(int x,int y)//判断是否越界以及是否走过{    if(vis[x][y]==0&&x>0&&x<=r&&y>0&&y<=c&&flag==0)    return true;    else    return false;}void dfs(int x,int y,int step){    p[step].s=x;    p[step].s1='A'+y-1; //换成字符    if(step==r*c)  //结束    {        flag=1;        return;    }    for(int i=0;i<8;i++)    {        int xx=x+dir[i][0];        int yy=y+dir[i][1];        if(judge(xx,yy))        {            vis[xx][yy]=1;            dfs(xx,yy,step+1);            vis[xx][yy]=0;        }    }}int main(){    cas=0;    cin>>t;    while(t--)    {      cas++;      cin>>r>>c;      p[1].s1='A';      p[1].s=1;      flag=0;      memset(vis,0,sizeof(vis));      vis[1][1]=1;      dfs(1,1,1);      if(flag)        print();      else      {          cout<<"Scenario #"<<cas<<":"<<endl;          cout<<"impossible"<<endl;          if(t!=0)            cout<<endl;      }    }    return 0;}

心得：

不是很难得题目，但需要细心仔细，多总结！