寒假训练 A - A Knight's Journey 搜索
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
dfs
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
int p,q;
int flag=0;
bool vis[100][100];
int path[100][100];
int tx[8]={-1,1,-2,2,-2,2,-1,1};
int ty[8]={-2,-2,-1,-1,1,1,2,2};
bool judge(int x,int y)
{
if(x>=1&&y>=1&&x<=p&&y<=q&&!flag&&!vis[x][y]) return true;
return false;
}
void dfs(int x,int y,int step)//这里不知道你们有没有问题,反正我碰到了,就是,这里的x和y,与坐标系不一样
{
path[step][0]=x;//x代表行
path[step][1]=y;//y代表列
if(step==p*q)//直角坐标系中x虽然是横轴,但是x的改变则是列的变化
{
flag=1;
return ;
}
for(int i=0;i<8;i++)
{
int sx=x+tx[i];
int sy=y+ty[i];
if(judge(sx,sy))
{
vis[sx][sy]=1;
dfs(sx,sy,step+1);
vis[sx][sy]=0;
}
}
} int main()
{
int n,c=0;
cin>>n;
while(n--)
{
scanf("%d%d",&p,&q);
printf("Scenario #%d:\n",++c);
memset(vis,0,sizeof(vis));
flag=0;
vis[1][1]=1;
dfs(1,1,1);
if(flag==1)
{
for(int i=1;i<=p*q;i++)
{
printf("%c%d",path[i][1]-1+'A',path[i][0]);
}
printf("\n");
}
else printf("impossible\n");
if(n!=0) printf("\n");
}
return 0;
}
ow many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .
Output
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 题目大意:就是给你p行q列,求马是否可以走完,可以求出路径,不可以输出-1
意思很明确,不过毕竟是英文题,有点难读
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