POJ 3083 Children of the Candy Corn（顺时针DFS+逆时针DFS+BFS）

  1 /*
  2 ** POJ 3083 Children of the Candy Corn
  3 ** Created by Rayn @@ 2014/05/09
  4 ** 这道题重点是left和right的DFS的方向问题，方向还和游客
  5 ** 当时朝向有关，BFS就是很普通的最短路搜索
  6 ** left搜索要遵循左上右下的顺时针搜索顺序
  7 ** right搜索要遵循右上左下的逆时针搜索顺序
  8 ** 朝向标记 ：左0 上1 右2 下3
  9 */
 10 #include <cstdio>
 11 #include <cstring>
 12 #include <queue>
 13 using namespace std;
 14 const int MAX = 50;
 15 struct pos{
 16     int x, y;
 17     int step;
 18 };
 19 int left, right;
 20 int w, h, sign[MAX][MAX];
 21 char maze[MAX][MAX];
 22 int dir[][2] = {{-1,0},{0,1},{1,0},{0,-1}}; /*0-上,1-右,2-下,3-左*/
 23 
 24 void leftDFS(int x, int y, int face)
 25 {
 26     int to[4] = {3,0,1,2};
 27 
 28     if(maze[x][y] == ‘E‘)
 29     {
 30         printf("%d ", left);
 31         return ;
 32     }
 33     left++;
 34     for(int i=0; i<4; ++i)
 35     {
 36         int nowface = (face + to[i]) % 4;
 37         int tx = x + dir[nowface][0];
 38         int ty = y + dir[nowface][1];
 39         if(tx>=0 && ty>=0 && tx<h && ty<w && maze[tx][ty] != ‘#‘)
 40         {
 41             leftDFS(tx, ty, nowface);
 42             return ;
 43         }
 44     }
 45 }
 46 void rightDFS(int x, int y, int face)
 47 {
 48     int to[4] = {1,0,3,2};
 49 
 50     if(maze[x][y] == ‘E‘)
 51     {
 52         printf("%d ", right);
 53         return ;
 54     }
 55     right++;
 56     for(int i=0; i<4; ++i)
 57     {
 58         int nowface = (face + to[i]) % 4;
 59         int tx = x + dir[nowface][0];
 60         int ty = y + dir[nowface][1];
 61         if(tx>=0 && ty>=0 && tx<h && ty<w && maze[tx][ty] != ‘#‘)
 62         {
 63             rightDFS(tx, ty, nowface);
 64             return ;
 65         }
 66     }
 67 }
 68 void BFS(int x, int y)
 69 {
 70     queue<pos> q;
 71     while(!q.empty())
 72         q.pop();
 73 
 74     pos start;
 75     start.x = x; start.y = y; start.step = 1;
 76     q.push(start);
 77     sign[x][y] = 1;
 78 
 79     while(!q.empty())
 80     {
 81         pos now = q.front();
 82         if(maze[now.x][now.y] == ‘E‘)
 83         {
 84             printf("%d\n", now.step);
 85             return ;
 86         }
 87         q.pop();
 88         for(int i=0; i<4; ++i)
 89         {
 90             pos tmp = now;
 91             tmp.x = now.x + dir[i][0];
 92             tmp.y = now.y + dir[i][1];
 93             tmp.step = now.step + 1;
 94             if(tmp.x>=0 && tmp.y>=0 && tmp.x<h && tmp.y<w && maze[tmp.x][tmp.y] != ‘#‘ && !sign[tmp.x][tmp.y])
 95             {
 96                 sign[tmp.x][tmp.y] = 1;
 97                 q.push(tmp);
 98             }
 99         }
100     }
101 }
102 int main()
103 {
104     int cases;
105     int sx, sy, face;
106 
107     scanf("%d", &cases);
108     while(cases--)
109     {
110         memset(sign, 0, sizeof(sign));
111 
112         scanf("%d%d%*c", &w, &h);
113         for(int i=0; i<h; ++i)
114         {
115             scanf("%s", maze[i]);
116             for(int j=0; j<w; ++j)
117             {
118                 if(maze[i][j] == ‘S‘)
119                 {
120                     sx = i;
121                     sy = j;
122                 }
123             }
124         }
125         //确定初始方位
126         if(sx == 0)
127             face = 2;
128         else if(sx == h-1)
129             face = 0;
130         if(sy == 0)
131             face = 1;
132         else if(sy == w-1)
133             face = 3;
134 
135         left = 1;
136         leftDFS(sx, sy, face);
137         right = 1;
138         rightDFS(sx, sy, face);
139         BFS(sx, sy);
140     }
141     return 0;
142 }