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Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

经典的DFS，大概题意就是让骑士走遍棋盘的所有格子。一路走到黑，看这条路能不能使骑士把格子都走完。

 #include<cstdio>
 #include<string.h>
 using namespace std;
 int vis[][];
 char str[];
 int s[][]={-,-,-,,-,-,-,,,-,,,,-,,};//这种处理方式很常用的，要记住，即骑士当前所在位置的周围所能走到的位置
 int p,q;
 int dfs(int x,int y,int sum,int cnt)
 {
     if(sum==p*q) return ;//sum是记录走过的格子
     int x1,y1;
     for(int i=;i<;i++)
     {
         x1=x+s[i][];
         y1=y+s[i][];
         if(x1>=&&x1<q&&y1>=&&y1<p&&!vis[x1][y1])//边界条件及判断是否访问过吗
         {
             vis[x1][y1]=;
             str[cnt+]=x1+'A';//开一个str数组，用来记录
             str[cnt+]=y1+'';
             if(dfs(x1,y1,sum+,cnt+))//记得这里要写成sum+1,cnt+2
                 return ;
             vis[x1][y1]=;//回溯
         }
     }
     return ;
 }
 int main()
 {
     int t;
     scanf("%d",&t);
     for(int i=;i<=t;i++)
     {
         scanf("%d %d",&p,&q);
         memset(vis,,sizeof(vis));
         memset(str,,sizeof(str));
         str[]='A';
         str[]='';
         vis[][]=;
         if(dfs(,,,)){
             printf("Scenario #%d:\n",i);
             for(int j=;j<strlen(str);j++)
                 printf("%c",str[j]);
             printf("\n\n");//记得每组数据输出后，有个空行
         }
         else{
             printf("Scenario #%d:\n",i);
             printf("impossible\n\n");
         }
     }
     return ;
 }