Description
Background
The knight is getting bored of seeing the same black and white squares againand again and has decided to make a journey
around the world. Whenever a knight moves, it is two squares in one directionand one square perpendicular to this. The world of a knight is the chessboardhe is living on. Our knight lives on a chessboard that has a smaller area thana regular 8 * 8 board, but
it is still rectangular. Can you help thisadventurous knight to make travel plans?
Problem
Find a path such that the knight visits every square once. The knight can startand end on any square of the board.
Input
The input begins with a positive integer n in the firstline. The following lines contain n test cases. Each test case consists of asingle 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 differentsquare numbers 1, . . . , p exist, q describes how many different square lettersexist. These are the first q letters of the Latin alphabet: A, . . .
Output
The output for every scenario begins with a linecontaining "Scenario #i:", where i is the number of the scenariostarting at 1. Then print a single line containing the lexicographically firstpath that visits all squares
of the chessboard with knight moves followed by anempty line. The path should be given on a single line by concatenating thenames of the visited squares. Each square name consists of a capital letterfollowed 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
給定一個p*q的棋盤，問騎士能不能走遍棋盤上的所有格子，每個格只訪問一次，可以從任意位置開始。
一道簡單的DFS，騎士必須按日”字行走。從棋盤上每個格作爲起點開始深搜，直到找到答案，或者所有情況都搜索到而沒有答案。
是模板題吧（大概）。不過自己當時做的時候，記錄每一步的方法太蠢，後來借鑑其他大牛的方法，效果好很多。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
using namespace std;
int p,q,path[30][2];
bool vis[30][30];
int dq[]={-2,-2,-1,-1,1,1,2,2};
int dp[]={-1,1,-2,2,-2,2,-1,1};
bool flag;
bool judge(int a,int b)
{
if (a>0&&a<=b)
return 1;
else
return 0;
}
void DFS(int x,int y,int cur)
{
if (cur==p*q)
{
flag=1;
return ;
}
else
{
for (int i=0;i<8;i++)
{
int tp,tq;
tp=x+dp[i]; tq=y+dq[i];
if (!vis[tp][tq]&&judge(tq,q)&&judge(tp,p))
{
path[cur+1][0]=tp; path[cur+1][1]=tq;
vis[tp][tq]=1;
DFS(tp,tq,cur+1);
if (flag)
return ;
vis[tp][tq]=0;
}
}
}
}
int main()
{
int T,n=0;
cin>>T;
while (T--)
{
if (n++)
printf("\n");
cin>>p>>q;
memset(vis,0,sizeof vis);
memset(path,-1,sizeof path);
flag=0;
for (int i=1;i<=p;i++)
{
for (int j=1;j<=q;j++)
{
memset(path,-1,sizeof path);
path[1][0]=1; path[1][1]=1;
vis[i][j]=1;
DFS(i,j,1);
if (flag)
break;
vis[i][j]=0;
}
if (flag)
break;
}
printf("Scenario #%d:\n",n);
if (flag)
{
for (int i=1;i<=p*q;i++)
{
printf("%c%d",'A'+path[i][1]-1,path[i][0]);
}
printf("\n");
}
else
{
printf("impossible\n");
}
}
return 0;
}