标签：ons include world lang put oid end pac code
题目：
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
Output
Sample Input
3
1 1
2 3
4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
题意：
给你一个p*q的棋盘，跳马在上面的任意一格开始移动，只能走‘日’字，问你能不能经过棋盘上面所有的格子；（百度的题意，明明是说经过所有的格子，有道硬是翻译
成了“找到一条这样的路，骑士每一次都要去一次”），输出要按照字典顺序输出
分析：
深度优先搜索，要经过所有的格子，那就肯定经过（1，1），所以就可以从（1，1）开始搜索；
AC代码：
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int t,p,q,flag;
int a[30][30];
int step[30][30];
int f[8][2]={{1,-2},{-1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
void dfs(int x,int y,int z)
{
step[z][1]=x;
step[z][2]=y;
if (z==p*q)
{
flag=1;
return ;
}
for (int i=0;i<8;i++)
{
int xi=x+f[i][0];
int yi=y+f[i][1];
if (xi>=1&&xi<=p&&yi>=1&&yi<=q&&!a[xi][yi]&&!flag)
{
a[xi][yi]=1;
dfs(xi,yi,z+1);
a[xi][yi]=0;
}
}
}
int main()
{
cin>>t;
for (int i=1;i<=t;i++)
{
flag=0;
scanf("%d%d",&p,&q);
memset(a,0,sizeof(a));
memset(step,0,sizeof(step));
a[1][1]=1;
dfs(1,1,1);
printf("Scenario #%d:\n",i);
if (flag==1)
{
for (int j=1;j<=p*q;j++)
printf("%c%d",step[j][2]+‘A‘-1,step[j][1]);
printf("\n");
}
else
printf("impossible\n");
if (i!=t)
printf("\n");
}
return 0;
}
标签：ons include world lang put oid end pac code
原文地址：http://www.cnblogs.com/lisijie/p/7243871.html
