zoj.3865.Superbot(bfs + 多维dp)

Superbot

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Superbot is an interesting game which you need to control the robot on an N*M grid map.

As you see, it's just a simple game: there is a control panel with four direction left (1^st position), right (2^nd), up (3^rd) and down (4^th). For each second, you can do exact one of the following operations:

• Move the cursor to left or right for one position. If the cursor is on the 1^st position and moves to left, it will move to 4^th position; vice versa.
• Press the button. It will make the robot move in the specific direction.
• Drink a cup of hot coffee and relax. (Do nothing)

However, it's too easy to play. So there is a little trick: Every P seconds the panel will rotate its buttons right. More specifically, the 1^st position moves to the 2^nd position; the 2^nd moves to 3^rd; 3^rd moves to 4^th and 4^th moves to 1^st. The rotating starts at the beginning of the second.

Please calculate the minimum time that the robot can get the diamond on the map.

At the beginning, the buttons on the panel are "left", "right", "up", "down" respectively from left to right as the picture above, and the cursor is pointing to "left".

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains three integers N, M (2 <= N, M <= 10) and P (1 <= P <= 50), which represent the height of the map, the width of the map and the period that the panel changes, respectively.

The following lines of input contains N lines with M chars for each line. In the map, "." means the empty cell, "*" means the trap which the robot cannot get in, "@" means the initial position of the robot and "$" means the diamond. There is exact one robot and one diamond on the map.

Output

For each test case, output minimum time that the robot can get the diamond. Output "YouBadbad" (without quotes) if it's impossible to get the diamond.

Sample Input

4
3 4 50
@...
***.
$...
5 5 2
.....
..@..
.*...
$.*..
.....
2 3 1
*.@
$.*
5 5 2
*****
..@..
*****
$....
.....

Sample Output

12
4
4
YouBadbad

Hint

For the first example: 
0s: start
1s: cursor move right (cursor is at "right")
2s: press button (robot move right)
3s: press button (robot move right)
4s: press button (robot move right)
5s: cursor move right (cursor is at "up")
6s: cursor move right (cursor is at "down")
7s: press button (robot move down)
8s: press button (robot move down)
9s: cursor move right (cursor is at "left")
10s: press button (robot move left)
11s: press button (robot move left)
12s: press button (robot move left)

For the second example:
0s: start
1s: press button (robot move left)
2s: press button (robot move left)
--- panel rotated ---
3s: press button (robot move down, without changing cursor)
4s: press button (robot move down)

For the third example:
0s: start
1s: press button (robot move left)
--- panel rotated ---
2s: press button (robot move down)
--- panel rotated ---
3s: cursor move left (cursor is at "right")
--- panel rotated ---
4s: press button (robot move left)

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Author: DAI, Longao
Source: The 15th Zhejiang University Programming Contest

 #include<stdio.h>
 #include<queue>
 #include<string.h>
 #include<math.h>
 #include<algorithm>
 struct node
 {
     int x , y , pan , ti ;
     node () {}
     node (int x , int y , int pan , int ti) : x (x) , y (y) , pan (pan) , ti (ti) { }
 };

 int move[][] ;
 int n , m , p ;
 char  map[][] ;
 bool dp[][][][] ;

 void rotate (int x)
 {
     int a0[][] = {{ , -} , { , } , {- , } , { , }  } ;
     int a1[][] = {{ , } , { , -} , { , } , {- , }  } ;
     int a2[][] = {{- , } , { , } , { , -} , { , }  } ;
     int a3[][] = {{ , } , {- , } , { , } , { , -}  } ;
     if (x == )
     for (int i =  ; i <  ; i++) {
         for (int j =  ; j <  ; j++) {
             move[i][j] = a0[i][j] ;
         }
     }
     else if (x == )
         for (int i =  ; i <  ; i++) {
         for (int j =  ; j <  ; j++) {
             move[i][j] = a1[i][j] ;
         }
     }
     else if (x == )
         for (int i =  ; i <  ; i++) {
         for (int j =  ; j <  ; j++) {
             move[i][j] = a2[i][j] ;
         }
     }
     else
         for (int i =  ; i <  ; i++) {
         for (int j =  ; j <  ; j++) {
             move[i][j] = a3[i][j] ;
         }
     }
 }

 void bfs (int sx , int sy)
 {
     node ans , tmp ;
     ans = node (sx , sy ,  , ) ;
     std::queue <node> q ;
     while (!q.empty ()) q.pop () ;
     q.push (ans) ;
     dp[sx][sy][][] =  ;
     while (!q.empty ()) {
         ans = q.front () ; q.pop () ;
      //   printf ("s   (%d,%d)(%d)(%d)\n" , ans.x , ans.y , ans.pan , ans.ti) ;
         for (int i =  ; i <  ; i++) {
             tmp ;
             int time = ans.ti + (fabs (i - ans.pan) ==  ?  : fabs(i - ans.pan)) ;
             rotate ((time/p)%) ;
             tmp.x = ans.x + move[i][] ; tmp.y = ans.y + move[i][] ;
             tmp.pan = i ;
             if (tmp.x <  || tmp.y <   || tmp.x >= n || tmp.y >= m) continue ;
             if (map[tmp.x][tmp.y] == '*') continue ;
             tmp.ti =  + time ;
             if (tmp.ti > )  continue ;
             if (dp[tmp.x][tmp.y][tmp.pan][tmp.ti]) continue ;
           //  printf ("(%d,%d)(%d)(%d)\n" , tmp.x , tmp.y , tmp.pan , tmp.ti) ;
             dp[tmp.x][tmp.y][tmp.pan][tmp.ti] =  ;
             q.push (tmp) ;
         }
         if (ans.ti +  <= ) {
             ans.ti ++ ;
             if (!dp[ans.x][ans.y][ans.pan][ans.ti])
                 q.push (ans) ;
         }
     }
 }

 int main ()
 {
    // freopen ("a.txt" , "r" , stdin ) ;
     int T ;
     scanf ("%d" , &T) ;
     while (T--) {
         int x , y , ex , ey ;
         scanf ("%d%d%d" , &n , &m , &p) ;
         getchar () ;
         for (int i =  ; i < n ; i++) {
             gets (map[i]) ;
         }
         for (int i =  ; i < n ; i++) {
             for (int j =  ; j <m ; j++) {
                 if (map[i][j] == '@') {
                     x = i , y = j ;
                 }
                 if (map[i][j] == '$') {
                     ex = i , ey = j ;
                 }
             }
         }
         memset (dp ,  , sizeof(dp)) ;
         bfs (x , y) ;
         int tim =  ;
         for (int i =  ; i <  ; i++) {
             for (int j =  ; j <  ; j++) {
                 if (dp[ex][ey][i][j] ) {
                    tim = std::min (tim , j) ;
                 }
             }
         }
         if (tim == ) puts ("YouBadbad") ;
         else printf ("%d\n" , tim) ;
     }
     return  ;
 }

一直让我很伤脑筋的问题是如何令 “ 喝咖啡"这个行动加入到队列中,后来问学长：

最大情况为10 * 10 * 4 * 200 = 8 * 10^4 很显然暴力过吧，orz。

所以不能回头的状态也用4维标志。