poj 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28335		Accepted: 9675

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

解题思路：深搜+回溯，并记录路径；

#include<stdio.h>
#include<string.h>
#define N 30
int dx[8]={-2,-2,-1,-1,1,1,2,2};
int dy[8]={-1,1,-2,2,-2,2,-1,1};
int mark

,n,m,ff;
char ans[2*N],cur[N*2];
struct node
{
int x,y;      //x记录横坐标，y记录纵坐标
}f
;
int judge(int x,int y)
{
if(x>0&&x<=n&&y>0&&y<=m)
return 1;
return 0;
}
void dfs(int x,int y,int num)
{
int i,k,di,dj;
if(num==n*m)      //找到答案后要立即记录路径
{
for(k=0;k<n*m;k++)   //不能用i,换一个变量控制循环
{
cur[2*k]=(char)f[k].x-1+'A';    //把横纵坐标转换为字符型字母和数字
cur[2*k+1]=(char)f[k].y+'0';
}
cur[2*k]='\0';
if(ff)
{
if(strcmp(cur,ans)<0)       //取字典序最小的序列
strcpy(ans,cur);
}
else
{
ff=1;
strcpy(ans,cur);
}
return ;
}
for(i=0;i<8;i++)
{
di=x+dx[i];
dj=y+dy[i];
if(judge(di,dj)&&!mark[di][dj])
{
mark[di][dj]=1;
f[num].x=di;
f[num].y=dj;
dfs(di,dj,num+1);
mark[di][dj]=0;  //回溯
}
}
return ;
}
int main()
{
int i,j,k,T,cnt=1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&m,&n);
for(i=1;i<=n;i++)     //字母在前
{
for(j=1;j<=m;j++)         //数字在后
{
memset(mark,0,sizeof(mark));
mark[i][j]=1;
f[0].x=i;
f[0].y=j;
ff=0;
dfs(i,j,1);
if(ff)   //搜到的第一个答案保证是字典序最小的
break;
}
if(ff)
break;
}
printf("Scenario #%d:\n",cnt++);
if(!ff)
printf("impossible\n\n");
else
printf("%s\n\n",ans);
}

return 0;
}