A Knight's Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 41972 | Accepted: 14286 |
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, . . .
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
Source
TUD Programming Contest 2005, Darmstadt, Germany
题意：
象棋中骑士走日，给出一个p*q棋盘，问其实能否一次走遍所有的格子，按照字典序输出路径。
代码：
//基础dfs,用vector保存路径。
#include<iostream>
#include<vector>
#include<cstdio>
#include<cstring>
using namespace std;
int p,q,t;
const int diry[]={-,,-,,-,,-,};
const int dirx[]={-,-,-,-,,,,};
int sum;
bool vis[][];
vector<int>loadx;
vector<int>loady;
void dfs(int x,int y)
{
vis[x][y]=;
sum++;
loadx.push_back(x);
loady.push_back(y);
if(sum==p*q)
return;
for(int i=;i<;i++)
{ if(x+dirx[i]<=||x+dirx[i]>q||y+diry[i]<=||y+diry[i]>p)
continue;
if(vis[x+dirx[i]][y+diry[i]])
continue;
dfs(x+dirx[i],y+diry[i]);
if(sum==p*q)
return;
}
vis[x][y]=;
sum--;
loadx.pop_back();
loady.pop_back();
}
int main()
{
scanf("%d",&t);
for(int k=;k<=t;k++)
{
scanf("%d%d",&p,&q);
sum=;
memset(vis,,sizeof(vis));
while(!loadx.empty())
{
loadx.pop_back();
loady.pop_back();
}
for(int i=;i<=q;i++)
{
if(sum==p*q)
break;
for(int j=;j<=p;j++)
{
dfs(i,j);
if(sum==p*q)
break;
}
}
printf("Scenario #%d:\n",k);
if(sum==p*q)
{
for(int i=;i<loadx.size();i++)
{
printf("%c%d",loadx[i]+,loady[i]);
}
printf("\n\n");
}
else printf("impossible\n\n");
}
return ;
}