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
![](http://poj.org/images/2488_1.jpg)
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
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; } }