A Knight’s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 35564 Accepted: 12119
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
Source
TUD Programming Contest 2005, Darmstadt, Germany
一段时间没有做搜索,果然手生了很多,犯了各种各样的错误,要赶紧补回来.
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#pragma comment(linker,"/STACK:102400000")
#define WW freopen("output.txt","w",stdout)
struct node
{
int x;
int y;
} a[100];
int Dir[][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
bool vis[9][9];
int n,m;
int flag;
bool DFS(int x,int y,int site)
{
vis[x][y]=true;
a[site].x=x;
a[site].y=y;
if(site==n*m)
{
flag=true;
return true;
}
int Fx,Fy;
for(int i=0; i<8; i++)
{
Fx=x+Dir[i][0];
Fy=y+Dir[i][1];
if(Fx>=0&&Fx<n&&Fy>=0&&Fy<m&&!vis[Fx][Fy])
{
if(DFS(Fx,Fy,site+1))
{
return true;
}
}
}
vis[x][y]=false;
return false;
}
int main()
{
int T,w=1;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&n,&m);
flag=false;
memset(vis,false,sizeof(vis));
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(DFS(i,j,1))
{
flag=true;
break;
}
}
if(flag)
{
break;
}
}
printf("Scenario #%d:
",w++);
if(flag)
{
for(int i=1; i<=n*m; i++)
{
printf("%c%c",a[i].y+'A',a[i].x+1+'0');
}
printf("
");
}
else
{
printf("impossible
");
}
}
return 0;
}
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