poj_2488 A Knight's Journey

比较典型的深搜，注意1，最后的输出格式，我imposiable后面忘记endl了，结果PE了两次，有点可惜； 2.最后要以字典序输出

 1 #include<iostream>
 2 #include<string.h>
 3 using namespace std;
 4
 5 #define MAX 8
 6
 7 int p,q;
 8 int board[MAX][MAX];
 9 int steps[MAX*MAX];
10
11 int dir[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
12
13
14 bool DFS(int x,int y,int step) {
15
16  if(step == p * q)
17    return true;
18
19  bool flag = false;
20  int tempx,tempy;
21
22  for(int i=0;i<8;i++) {
23
24
25    tempx = x + dir[i][0];
26    tempy = y + dir[i][1];
27
28    if(!board[tempx][tempy] && tempx < p && tempy < q && tempx >= 0 && tempy >= 0) {
29      board[tempx][tempy] = 1;
30      if(DFS(tempx,tempy,step+1))
31      {
32       steps[step-1] = i;
33       flag = true;
34       break;
35      }
36      else {
37      board[tempx][tempy] = 0;
38      }
39
40    }
41
42  }
43
44   return flag;
45
46 }
47
48
49
50
51 int main() {
52
53
54 int count;
55
56 cin >> count;
57 int num = count;
58 while(num--) {
59
60  cin>> p >> q;
61
62  memset(board,0,sizeof(int)*MAX*MAX);
63  memset(steps,0,sizeof(int)*MAX*MAX);
64  board[0][0] = 1;
65
66  cout<<"Scenario #"<<count-num<<":"<<endl;
67
68  if(DFS(0,0,1)) {
69    string s("A1");
70    int x_offset=0,y_offset=0;
71
72    for(int i=0;i<p*q-1;i++) {
73
74      x_offset += dir[steps[i]][0];
75      y_offset += dir[steps[i]][1];
76
77      s.push_back((char)(‘A‘+ y_offset));
78      s.push_back((char)(‘1‘+ x_offset));
79    }
80
81    cout<<s<<endl<<endl;
82  }
83  else
84    cout<<"impossible"<<endl<<endl;
85
86 }
87
88
89 return 0;
90 }

poj_2488 A Knight's Journey