UVa (BFS) The Monocycle

题目不光要求要到达终点而且要求所走的步数为5的倍数，每个时刻有三个选择，前进，左转弯，右转弯。

所以在vis数组中新增加两个维度即可，vis[x][y][dir][color]表示在(x, y)格子方向朝dir与地面接触的扇形的颜色为color，这个状态是否到达过。

 1 #include <cstdio>

 2 #include <queue>

 3 #include <cstring>

 4 using namespace std;

 5 

 6 const int maxn = 30;

 7 char maze[maxn][maxn];

 8 bool vis[maxn][maxn][4][5];

 9 

10 struct Node

11 {

12     int x, y, d, t, step;

13     Node(int x=0, int y=0, int d=0, int t=0, int step=0):x(x), y(y), d(d), t(t), step(step) {}

14 }st, ed;

15 

16 int dx[] = { -1, 0, 1, 0 };

17 int dy[] = { 0, 1, 0, -1 };

18 

19 int row, col;

20 

21 inline bool in(int x, int y)

22 { return x >= 0 && x < row && y >= 0 && y < col; }

23 

24 int BFS()

25 {

26     memset(vis, false, sizeof(vis));

27     queue<Node> Q;

28     Q.push(st);

29     vis[st.x][st.y][0][0] = true;

30     while(!Q.empty())

31     {

32         Node now = Q.front(); Q.pop();

33         if(now.x == ed.x && now.y == ed.y && now.step % 5 == 0)

34             return now.t;

35 

36         int d = now.d;

37         int x = now.x + dx[d];

38         int y = now.y + dy[d];

39         int t = now.t + 1;

40         int step = now.step + 1;

41         if(in(x, y) && maze[x][y] != '#' && !vis[x][y][d][step%5])

42         {

43             Q.push(Node(x, y, d, t, step));

44             vis[x][y][d][step%5] = true;

45         }

46 

47         for(int i = 1; i <= 3; i += 2)

48         {

49             x = now.x; y = now.y;

50             d = (now.d + i) % 4;

51             t = now.t + 1;

52             step = now.step;

53             if(!vis[x][y][d][step%5])

54             {

55                 Q.push(Node(x, y, d, t, step));

56                 vis[x][y][d][step%5] = true;

57             }

58         }

59     }

60     return -1;

61 }

62 

63 int main()

64 {

65     //freopen("in.txt", "r", stdin);

66 

67     int kase = 0;

68     while(scanf("%d%d", &row, &col) == 2)

69     {

70         if(row == 0 && col == 0) break;

71 

72         if(kase++ > 0) puts("");

73         printf("Case #%d\n", kase);

74 

75         if(row == 0 && col == 0) break;

76         for(int i = 0; i < row; i++) scanf("%s", maze[i]);

77         for(int i = 0; i < row; i++)

78             for(int j = 0; j < col; j++)

79             {

80                 if(maze[i][j] == 'S') st = Node(i, j, 0, 0, 0);

81                 if(maze[i][j] == 'T') ed = Node(i, j);

82             }

83 

84         int ans = BFS();

85         if(ans >= 0) printf("minimum time = %d sec\n", ans);

86         else puts("destination not reachable");

87     }

88 

89     return 0;

90 }

代码君