POJ - 2488 A Knight's Journey (DFS)
2017-08-10 14:09
471 查看
Language: Default A Knight's Journey
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 Source TUD Programming Contest 2005, Darmstadt, Germany |
我认为DFS的一般思路:
1.明确下一个方向是啥,你要进行的下一个动作是要干什么。比如这道题就是,下一步就是马走日,8种走法,需要注意字典序的问题,所以,搜索的顺序要确定好,纵坐标从A到Z,数字横坐标从小到大。
2.我认为主要问题在于不会写代码,所以下面贴出的代码多多少少有些借鉴。
(1)有三件事需要判断,是否搜超界,点是否访问过,是否遍历了整个棋盘,所以可以用一个函数来判断一下我能不能走下一步。
(2)ans数组负责记录你的搜索路线,返回的条件是是否遍历满整个格子,但是要是遍历不满怎么办呢?上面函数的作用就体现出来了,先判断你这个方法是否可行,然后再去搜索,这样,万一出现2*3的格子避免无限递归的尴尬。
(3)写递归类的程序扣得太细容易糊涂,这里对for循环做出如下解释:
搜索八个方向,对于每个方向,都执行一次DFS函数,那么这个函数大致干了些什么?它的作用就是生成一个遍历的方案!
这里加了一个判断条件,就是万一走棋盘外面,或者出现来回走无限递归的情况,我们就不生成方案了。至于那个flag啥用?其实就是一个标记,其实就相当于,你判断完方案不可行了,但是你得发出这个信号。如果全遍历完了,flag才等于1,那个judge函数里的!flag,是指都遍历完了就不能再次DFS走下一步了,所以都遍历完了也在判断范围内。
这就说明,当flag=1时说明这个方案可行,由于咱们按字典序分的搜索方向,所以这就是答案了;
(4)写主函数的时候,对于每个方案都要清空VIS数组,把flag归0,从哪个点开始搜索就带入DFS函数哪个实参,这里由于要求字典序,我们要从A1遍历。刚开始走就是第一步,所以DFS(1,1,1)。
以上是我对参考代码的理解,如果我还是不会写代码,我就还像这么样解释,直到掌握代码该怎么写了。
我的参考过的代码如下(自己写的根本不叫DFS,乱七八糟,明白那个意思但是程序跑了根本出不来结果或者出现奇怪结果,所以不贴了):
#include<iostream> #include<stdio.h> #include<string.h> const int maxn=80; int vis[maxn][maxn]; int ans[maxn][2]; int s[10][3]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2} }; int sx,sy; int flag; bool judge(int x, int y) { if(x >= 1 && x <= sx && y >= 1 && y <= sy && !vis[x][y] && !flag) return true; return false; } void DFS(int x,int y,int step) { ans[step][0]=x; ans[step][1]=y; if(sx*sy==step) { flag=1; return; } for(int i=0;i<8;i++) { int dx=x+s[i][0]; int dy=y+s[i][1]; if(judge(dx,dy)) { vis[dx][dy]=1; DFS(dx,dy,step+1); vis[dx][dy]=0; } } } int main() { int t; scanf("%d",&t); for(int j=1;j<=t;j++) { scanf("%d%d",&sx,&sy); memset(vis,0,sizeof(vis)); flag=0; vis[1][1]=1; DFS(1,1,1); printf("Scenario #%d:\n",j); if(flag) { for(int i=1;i<=sx*sy;i++) { printf("%c%d",ans[i][1]-1+'A',ans[i][0]); } printf("\n\n"); } else printf("impossible\n\n"); } return 0; }
相关文章推荐
- POJ训练计划2488_A Knight's Journey(DFS+回溯)
- POJ2488 A Knight's Journey(DFS)
- 【dfs】POJ-2488-A Knight's Journey
- A Knight's Journey POJ - 2488 dfs水题
- POJ 2488 A Knight's Journey【DFS】
- POJ 2488 A Knight's Journey【DFS + 回溯应用】
- POJ 2488 A Knight's Journey (DFS)
- poj 2488 A Knight's Journey 【dfs】【字典序】【刷题计划】
- POJ 2488 A Knight's Journey (DFS深搜)
- poj 2488 A Knight's Journey (dfs)
- POJ-2488 A Knight's Journey (DFS)
- POJ 2488 A Knight's Journey (dfs)
- POJ 2488 A Knight's Journey(DFS+回溯)
- poj 2488 -- A Knight's Journey (DFS)
- POJ 2488 A Knight's Journey(DFS)
- poj 2488 A Knight's Journey DFS
- POJ 2488 A Knight's Journey【dfs】
- 搜索 -dfs-POJ-2488-A Knight's Journey
- POJ 2488 A Knight's Journey DFS
- [DFS]poj 2488 A Knight's Journey