- 总时间限制:
- 1000ms
- 内存限制:
- 65536kB
- 描述
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.- 输入
- 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, . . .
- 输出
- 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. - 样例输入
-
3
1 1
2 3
4 3
- 样例输出
-
Scenario #1:
A1
Scenario #2:
impossible
Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
- 来源
- TUD Programming Contest 2005, Darmstadt, Germany
- OJ:OpenJudge
- 题目大意
- 给出一个p行q列的国际棋盘，马可以从任意一个格子开始走，问马能否不重复的走完所有的棋盘。如果可以，输出按字典序排列最小的路径。打印路径时，列用大写字母表示（A表示第一列），行用阿拉伯数字表示（从1开始），先输出列，再输出行。（摘自http://blog.csdn.net/lyhvoyage/article/details/18355471）国际象棋的马的走法如图所示，马走的是日字，和中国象棋是一样，是跳着走，但是国际象的马没有蹩脚，图片上标注的圆是马可以走的地方。
- 分析
- 1、要你找一个不重复走完所有棋盘的路径，所以用DFS+回溯，尝试多条路径，如果该路径不能走完就回溯。
- 2、能找到走完的所有格子的路径，就输出字典序最小的，又因为可以在任意一个各子开始，所以肯定直接从A1开始走，然后把遍历的方向按字典序从小到大排，这样就能保证按此方案能走完所有棋盘的路径就是题目要求的路径。
- 参考代码
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct path{//记录路径
int r;
char c;
}p[900];
int r,c;//row,column
int flag=0;//表示是否找到要求的路径
///p*q大于等于1小于等于26,所以p,q也是大于等于小于等于26
int visited[30][30];//标记数组，避免重复
/*
按字典序从小到大的顺序排列方向，
就能保证遍历时是字典序最小的路径
*/
const int di[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
const int dj[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
void dfs(int si,int sj,int step);
int main()
{
int n,num=1;
scanf("%d",&n);
while(n){
scanf("%d%d",&r,&c);
memset(visited,0,sizeof(visited));
visited[1][1]=1;//A1点直接初始化
dfs(1,1,1);
printf("Scenario #%d:
",num);
if(flag==1){
for(int i=1;i<=r*c;i++){
printf("%c%d",p[i].c,p[i].r);
}
printf("
");
}
else{
printf("impossible
");
}
flag=0;
n--;
num++;
}
return 0;
}
void dfs(int si,int sj,int step){
//记录路径
p[step].r=si;
p[step].c='A'+sj-1;
//一共r*c个点所以能走r*c步就是能走完所有格子
if(step==r*c){
flag=1;
return;
}
//遍历该点的八个方向
for(int i=0;i<8;i++){
int sii=si+di[i];
int sjj=sj+dj[i];
//必须满足这些条件才能进入函数
if(sii>0&&sii<=r&&sjj>0&&sjj<=c&&visited[sii][sjj]==0&&flag==0){
//进入函数的就标记下来
visited[sii][sjj]=1;
dfs(sii,sjj,step+1);
//不能走完的路径会回溯回来，把这些走过但是不符合要求的点重新化0，让其他方向的路径试
visited[sii][sjj]=0;
}
}
}