A Knight's Journey
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 51 Accepted Submission(s) : 17
Problem 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
1 1
2 3
4 3
Sample Output
Scenario #1:
A1
Scenario #2:
impossible
Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
Source
PKU
第一次写深搜，感觉还不错
#include<iostream>
#include<cstring>
using namespace std;
int m,n;
int c; //已走过的步数
bool f; //标记是否找到答案了
int chess[30][30]={0};
int step[2][8]={{-1, 1,-2, 2,-2,2,-1,1}, //每一种可能的走法，注意要按字典序排列
{-2,-2,-1,-1, 1,1, 2,2}};
char ans1[64];
int ans2[64];
bool move(int i,int j,int k)
{
if(i+step[0][k]>=0&&i+step[0][k]<m&&j+step[1][k]>=0&&j+step[1][k]<n&&!chess[i+step[0][k]][j+step[1][k]])
{
chess[i+step[0][k]][j+step[1][k]]=1;
return true;
}
else
return false;
}
void DFS(int i,int j)
{ ans1[c]=j+'A';
ans2[c]=i+1; if(c==m*n) //找到结果，回退
{
f=true;
return;
}
for(int t=0;t<8;t++) //尝试每一种走法
{
if(move(i,j,t))
{
c++;
DFS(i+step[0][t],j+step[1][t]);
if(c==m*n) //如果找到结果就一直回退
return;
chess[i+step[0][t]][j+step[1][t]]=0; //还原
c--;
}
}
return;
} int main()
{
int T;
cin>>T;
int t=T;
while(T--)
{
int i,j,k;
cin>>m>>n;
for(j=0;j<n;j++)
{
for(i=0;i<m;i++)
{
c=1;
memset(chess,0,sizeof(chess));
chess[i][j]=1;
f=false;
DFS(i,j);
if(f)
{
cout<<"Scenario #"<<t-T<<":"<<endl;
for(k=1;k<=m*n;k++)
cout<<ans1[k]<<ans2[k];
cout<<endl;
break;
}
}
if(f)
break;
}
if(!f)
{
cout<<"Scenario #"<<t-T<<":"<<endl;
cout<<"impossible"<<endl;
}
cout<<endl;
}
}