A Knight’s Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 43300 | Accepted: 14697 |
Description
BackgroundThe 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
题意
马要走遍p*q的棋盘的所有格，它可以从任意点出发，问能否成功，若能，输出所能走的最小字典序的路径。
思路
简单dfs，判断马是否走遍了所有格我们可以用搜索的深度来判断，如果它等于棋盘格的数量，则已经走遍了所有，在dfs过程中，我们要保证走过的点不会被重复踏入，另外一个便是如何保证字典序最小，可以采用搜索的顺续来确定，具体顺续见move数组。
AC 代码
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include<algorithm>
#include <queue>
using namespace std;
bool visited[30][30],flag;
int move[8][2]= {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int p,q;
struct node
{
char x;
char y;
void init(int x,int y)
{
this->x=x;
this->y=y;
}
} sk[1005];
void dfs(int x,int y,int deep)
{
if(deep==p*q) //已经走遍了所有格
{
for(int i=0; i<deep; i++)
printf("%c%c",sk[i].x,sk[i].y);
printf("\n");
flag=true; //找到路径
return;
}
for(int i=0; i<8; i++)
{
int xi=move[i][0]+x;
int yi=move[i][1]+y;
if(xi<0||xi>=p||yi<0||yi>=q)continue; //棋盘外
if(visited[xi][yi]==false) //该位置为空
{
visited[xi][yi]=true;
sk[deep].init(yi+'A',xi+'1');
dfs(xi,yi,deep+1);
if(flag)return;
visited[xi][yi]=false;
}
}
}
int main()
{
int n;
scanf("%d",&n);
for(int ni=1; ni<=n; ni++)
{
printf("Scenario #%d:\n",ni);
scanf("%d%d",&p,&q);
flag=false; //当前未找到路径
memset(visited,false,sizeof(visited));
for(int j=0; j<q; j++) //先枚举列是为了保证字典序最小
for(int i=0; i<p; i++)
{
visited[i][j]=true;
sk[0].init(j+'A',i+'1');
dfs(i,j,1);
visited[i][j]=false;
}
if(!flag) //没有找到路径
printf("impossible\n");
if(ni!=n)puts("");
}
return 0;
}
