A Knight’s Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 41945 | Accepted: 14272 |
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, . . .
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.
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
题意：
给你一个棋盘，问你一匹马能否不重复遍历整个棋盘，马只能走日字形，如果能走，就按字典序输出最短路径，否则，输出impossible。
分析：
首先，如果能遍历整个棋盘，那么从哪里走都可以，所以一定能从A1走，所以，我们从A1开始搜。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int p, q, t, vis[30][30];
bool flag = 0;
int f[2][8] = {
{-2,-2, -1,-1, 1,1, 2,2},
{-1, 1, -2, 2,-2,2,-1,1}
};
struct qer
{
int x, y;
}ans[50];
void dfs(int x, int y, int step, int tt)
{
if(flag)
return ;
vis[x][y] = 1;
ans[step].x = x;
ans[step].y = y;
if(step == p*q && flag == 0)
{
flag = 1;
printf("Scenario #%d:\n", tt);
for(int i = 1;i <= step;i ++)
{
char c = ans[i].x - 1 + 'A';
cout<<c<<ans[i].y;
}
puts("");
puts("");
}
for (int i = 0;i < 8;i ++)
{
int nex = x + f[0][i];
int ney = y + f[1][i];
if (nex >= 1 && nex <= q && ney >= 1 && ney <= p && vis[nex][ney] != 1)
dfs(nex, ney, step + 1, tt);
}
vis[x][y] = 0;
}
int main()
{
cin>>t;
int tot = 0;
while(t--)
{
tot ++;
memset(vis, 0, sizeof(vis));
flag = 0;
scanf("%d%d", &p, &q);
dfs(1, 1, 1, tot);
if(!flag)
{
printf("Scenario #%d:\n", tot);
printf("impossible\n");
puts("");
}
}
return 0;
}