問題:
http://acm.pku.edu.cn/JudgeOnline/problem?id=3083思路:
對于求最短路徑，BFS即可解決，沒有什么難度
該題的難點在于如何求沿左、沿右走的問題
剛開始，完全不知道這是什么意思，無奈只能在網上看代碼，總結如下:
沿左策略的一般次序: left, up, right, down
沿右策略的一般次序: right, up, left, down
求解的關鍵是如何根據前一個方向以及一般次序來決定目前訪問上下左右四個方向的順序，例如:
對于沿左前進策略，如果前一個方向是right，那么訪問次序是up, right, down, left(與前一個方向相反的方向總是放在最后)
代碼:
1 #define MAX_LEN 42
2 #define QUEUE_LEN 1600
3 #define is_valid(x, y) (x>=0 && x<row && y>=0 && y<col)
4 char maze[MAX_LEN][MAX_LEN];
5 int visited[MAX_LEN][MAX_LEN];
6 int row, col;
7 int start_x, start_y;
8 /* left, up, right, down */
9 const int dx[] = {0, -1, 0, 1};
10 const int dy[] = {-1, 0, 1, 0};
11 /* right, up, left, down */
12 const int dx_right[] = {0, -1, 0, 1};
13 const int dy_right[] = {1, 0, -1, 0};
14 int lcount, rcount;
15 int head, tail;
16 struct EACH {
17 int x, y;
18 int mv;
19 } queue[QUEUE_LEN];
20
21
22 void
23 init()
24 {
25 int i;
26 char *p;
27 memset(visited, 0, sizeof(visited));
28 head = -1;
29 tail = 0;
30 lcount = rcount = 0;
31 scanf("%d %d", &col, &row);
32 for(i=0; i<row; i++) {
33 scanf("%s", maze[i]);
34 if((p=strchr(maze[i], 'S')) != NULL) {
35 start_x = i;
36 start_y = p-maze[i];
37 }
38 }
39 }
40
41 /*
42 * dir: previous direction
43 * switch(dir):
44 * case(right): up right down left (order)
45 * case(up): left up right down
46 * case(left): down left up right
47 * case(down): right down left up
48 */
49 void
50 dfs_left(int x, int y, int dir)
51 {
52 int i, tx, ty;
53 ++lcount;
54 if(maze[x][y] == 'E') {
55 return;
56 }
57 dir = (dir+3)%4;
58 for(i=0; i<4; i++) {
59 tx = x + dx[(dir+i)%4];
60 ty = y + dy[(dir+i)%4];
61 if(is_valid(tx, ty) && maze[tx][ty]!='#') {
62 dfs_left(tx, ty, (dir+i)%4);
63 break;
64 }
65 }
66 }
67
68 void
69 dfs_right(int x, int y, int dir)
70 {
71 int i, tx, ty;
72 ++rcount;
73 if(maze[x][y] == 'E')
74 return;
75 dir = (dir+3)%4;
76 for(i=0; i<4; i++) {
77 tx = x + dx_right[(dir+i)%4];
78 ty = y + dy_right[(dir+i)%4];
79 if(is_valid(tx, ty) && maze[tx][ty]!='#') {
80 dfs_right(tx, ty, (dir+i)%4);
81 break;
82 }
83 }
84 }
85
86 int
87 bfs()
88 {
89 int i, cx, cy, tx, ty, cmv;
90 memset(visited, 0, sizeof(visited));
91 queue[tail].x = start_x;
92 queue[tail].y = start_y;
93 queue[tail].mv = 1;
94 visited[start_x][start_y] = 1;
95 while(head < tail) {
96 ++head;
97 cx = queue[head].x;
98 cy = queue[head].y;
99 cmv = queue[head].mv;
100 if(maze[cx][cy] == 'E')
101 return cmv;
102 for(i=0; i<4; i++) {
103 tx = cx + dx[i];
104 ty = cy + dy[i];
105 if(is_valid(tx, ty) && !visited[tx][ty] && maze[tx][ty]!='#') {
106 ++tail;
107 queue[tail].x = tx;
108 queue[tail].y = ty;
109 queue[tail].mv = cmv+1;
110 visited[tx][ty] = 1;
111 }
112 }
113 }
114 }