A Knight's Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 28630 | Accepted: 9794 |
Description
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, . . .
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.
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 【题目背景】
我们知道，在国际象棋中, 骑士的移动路线是 L 型的, (在水平和垂直两个方向上, 一个方向上走两格, 另一个方向上走一格). 因此, 在一个空棋盘中间的方格上, 骑士
可以有 8 种不同的移动方式.
这题的问题是：如果马从[0,0]出发，能否将棋盘中的每一个格子走遍并且保证每个格子直走一次，如果可以的话，按照字典序输出路径。 【题目分析】
这题其实就是一个简单的DFS，难点在于按照字典序来输出路径，这就需要自己画图分析了。 下面是我的AC代码：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
bool vis[][]; //标记是否走过
int path[][]; //记录走过路径
bool Find;
int a,b;
int dir[][]={-,-, -,, -,-, -,, ,-, ,, ,-, ,}; //实际是一个闭合的环
void dfs(int i,int j,int k)
{
if(a*b==k)
{
for(int i=;i<k;i++)
{
printf("%c%d",path[i][]+'A',path[i][]+);
}
puts("");
Find=;
}
else
{
for(int x=;x<;x++)
{
int n=i+dir[x][];
int m=j+dir[x][];
if(n>=&&n<b&&m>=&&m<a&&!vis[n][m]&&!Find)
{
vis[n][m]=;
path[k][]=n;
path[k][]=m;
dfs(n,m,k+);
vis[n][m]=;
}
}
}
}
int main()
{
int kase=;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&a,&b);
Find=;
memset(vis,,sizeof(vis));
vis[][]=;
path[][]=;
path[][]=;
printf("Scenario #%d:\n",kase++);
dfs(,,);
if(!Find)
printf("impossible\n");
puts("");
}
return ;
}