poj 3083 

  1 #include <iostream>
  2 #include <limits>
  3 #include <queue>
  4 using namespace std;
  5 
  6 typedef pair<int, int> P;
  7 
  8 char maze[41][41]; 
  9 int dist[41][41];
 10 int spDirX[4] = {1, 0, -1, 0};
 11 int spDirY[4] = {0, 1, 0, -1};
 12 int n;
 13 int w, h;
 14 int startX, startY;
 15 int endX, endY;
 16 const int INF = numeric_limits<int>::max();
 17 int sp, lo, ro;
 18 int dx[4] = {
 19         0, 1, 0, -1
 20 };
 21 int dy[4] = {
 22         -1, 0, 1, 0    
 23 };
 24 void leftOrder(int x, int y, int sum, int dir)
 25 {
 26     //ÖÕÖ¹ 
 27     if(x == endX && y == endY)
 28     {
 29         lo = sum;
 30         return ;
 31     }
 32     //µÚÒ»¸öÔªËØÈ·¶¨·½Ïò
 33     if(sum == 1)
 34     {
 35         for(int i = 0; i < 4; ++i)
 36         {
 37             int nx = x + dx[i];
 38             int ny = y + dy[i];
 39             if(nx >= 0 && nx < w && ny >= 0 && ny < h && maze[ny][nx] == '.')
 40             {
 41                     leftOrder(nx, ny , sum+1, i);
 42                     break;
 43             }
 44             
 45         }
 46     }
 47     else
 48     {
 49         //Ò»°ãÇé¿ö 
 50         dir = (dir+3) % 4;
 51         for(int i = 0; i < 4; ++i)
 52         {
 53             int nx = x + dx[(dir+i)%4];
 54             int ny = y + dy[(dir+i)%4];
 55             if(nx >= 0 && nx < w && ny >= 0 && ny < h && maze[ny][nx] != '#')
 56             {
 57                 leftOrder(nx, ny , sum+1, (dir+i)%4);
 58                 break;
 59             }        
 60         }    
 61     }
 62     
 63 }
 64 
 65 void rightOrder(int x, int y, int sum, int dir)
 66 {
 67         //ÖÕÖ¹ 
 68     if(x == endX && y == endY)
 69     {
 70         ro = sum;
 71         return ;
 72     }
 73     //µÚÒ»¸öÔªËØÈ·¶¨·½Ïò
 74     if(sum == 1)
 75     {
 76         for(int i = 0; i < 4; ++i)
 77         {
 78             int nx = x + dx[i];
 79             int ny = y + dy[i];
 80             if(nx >= 0 && nx < w && ny >= 0 && ny < h && maze[ny][nx] == '.')
 81             {
 82                     rightOrder(nx, ny , sum+1, i);
 83                     break;
 84             }
 85             
 86         }
 87     }
 88     else
 89     {
 90         //Ò»°ãÇé¿ö 
 91         dir = (dir+1) % 4;
 92         for(int i = 0; i < 4; ++i)
 93         {
 94             int nx = x + dx[(dir-i+4)%4];
 95             int ny = y + dy[(dir-i+4)%4];
 96             if(nx >= 0 && nx < w && ny >= 0 && ny < h && maze[ny][nx] != '#')
 97             {
 98                 rightOrder(nx, ny , sum+1, (dir-i+4)%4);
 99                 break;
100             }        
101         }    
102     }
103     
104 } 
105 
106 int shortPath()
107 {
108     queue<P> que; 
109     dist[startY][startX] = 1;
110     que.push(P(startX, startY));
111     while(que.size())
112     {
113         P temp = que.front();
114         que.pop();
115         if(temp.first == endX && temp.second == endY)
116             break;
117         for(int i = 0; i < 4; ++i)
118         {
119             int nx = temp.first + spDirX[i];
120             int ny = temp.second + spDirY[i];
121             if(nx >= 0 && ny >= 0 && nx < w && ny < h && dist[ny][nx] == INF && maze[ny][nx] != '#')
122             {
123                 que.push(P(nx, ny));
124                 dist[ny][nx] = dist[temp.second][temp.first] + 1;
125             }
126         }        
127     }
128     sp = dist[endY][endX];
129 }
130 
131 
132 int main()
133 {
134     cin >> n;
135     for(int i = 0; i < n; ++i)
136     {
137         cin >> w >> h;
138         for(int j = 0; j < h; ++j)
139         {
140             for(int k = 0; k < w; ++k)
141             {
142                 cin >> maze[j][k];
143                 dist[j][k] = INF;
144                 if(maze[j][k] == 'S')
145                 {
146                     startX = k;
147                     startY = j;
148                 }
149                 if(maze[j][k] == 'E')
150                 {
151                     endX = k;
152                     endY = j;
153                 }    
154             }
155         }
156         leftOrder(startX, startY, 1, 0);
157         rightOrder(startX, startY, 1, 0);
158         shortPath();    
159         cout << lo << " " << ro << " " << sp << endl;    
160     }
161     return 0;
162 }
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posted @ 2014-11-05 00:07  多解方程式  阅读(281)  评论(0)    收藏  举报