A Knight's Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
|---|---|---|
| Total Submissions: 35837 | Accepted: 12216 |
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
测试数据：
3 7
Scenario #3:
A1B3D2F1G3E2G1F3E1G2E3C2A3B1C3A2C1D3B2D1F2
3 8
Scenario #4:
A1B3C1A2C3D1B2D3E1G2E3C2A3B1D2F1H2F3G1E2G3H1F2H3
题目大意：
人走“日”字型，判断能不能走完全部的方格，若能并输出走的顺序。
思路：就是一个深搜。不过个人认为有个难点：如何记录下所有经过的路径，并DFS返回时该怎么处理。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include<algorithm>
#include<math.h>
using namespace std;
int vis[80][80];
int tx[80],ty[80];
int n,m;
int cx[]= {-1,1,-2,2,-2,2,-1,1};
int cy[]= {-2,-2,-1,-1,1,1,2,2};//此顺次是固定的从下道上，从左到右的次序，并且是对称的
int bj;
void dfs(int s1,int s2,int ans)
{
tx[ans]=s1;//一定是刚进DFS就存下坐标，否则在return后最后一个路径下标就没法存起来
ty[ans]=s2;
if(ans==n*m)//搜索完成的标志
{
bj=1;
return ;
}
for(int i=0; i<8; i++)
{
int x=cx[i]+s1;
int y=cy[i]+s2;
if(x>=1&&y>=1&&x<=n&&y<=m&&!vis[x][y]&&!bj)//注意不要扔掉!bj，如果查找完毕后函数饭回的话，没有此句依然会进行
{
vis[x][y]=1;
dfs(x,y,ans+1);
vis[x][y]=0;
}
}
}
int main()
{
int cla;
scanf("%d",&cla);
for(int gr=1; gr<=cla; gr++)
{
bj=0;
memset(vis,0,sizeof(vis));
memset(tx,0,sizeof(tx));
memset(ty,0,sizeof(ty));
printf("Scenario #%d:\n",gr);
scanf("%d%d",&n,&m);
vis[1][1]=1;
dfs(1,1,1);//便收索边进行步数的计算
if(bj)
{
for(int i=1; i<=n*m; i++)
{
printf("%c%d",ty[i]-1+'A',tx[i]);
}
}
else
printf("impossible");
printf("\n");
if(gr!=cla)
printf("\n");
}
return 0;
}
