ZOJ 3865 Superbot(优先队列--模板)

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5477

主要思路：1.从一个点(cur)到它相邻的点(next)，所需要的时间数(t)其实是固定的，而且这个移动过程后，到达next时，相应的方向也是固定的，找到求t的办法就好了。

　　　　   　 2.到达一个未到达的点可能有多条路，优先队列取时间最短的路，则答案最优

题目:

Superbot

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^rdmoves 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)

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

代码：

 #include "cstdio"
 #include "cstring"
 #include "queue"
 using namespace std;

 #define N 15
 #define INF 0x3333333
 char map[N][N];
 int n,m,p;
 int mark[N][N][]; //到达每个点4个方向上的最短时间
 int dir[][] = {{,-},/*0:left*/{,},/*1:right*/{-,}/*2:up*/,{,}/*3:down*/};

 typedef struct node
 {
     int x,y;  //coordinate
     int t;  //time
     int di; //direction  0:left 1:right 2:up 3:down
     bool operator < (const node &b) const{
         return t > b.t;
     }
     /*
     friend bool operator<(node a,node b)
     {
         return a.di > b.di;
     }
     */
 } Point;

 Point st,en;

 void Init()  //标记数组初始化
 {
     for(int i=; i<=n; i++)
     {
         for(int j=; j<=m; j++)
         {
             for(int k=; k<; k++)
                 mark[i][j][k] = INF;
         }
     }
 }

 void Init_st_en()
 {
     for(int i=; i<=n; i++)
     {
         for(int j=; j<=m; j++)
         {
             if(map[i][j]=='@')
                 st.x=i,st.y=j,st.t=,st.di=;
             if(map[i][j]=='$')
                 en.x=i,en.y=j;
         }
     }
 }

 int DFS();

 int main()
 {
     int i,T;
     int ans;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d %d %d",&n,&m,&p);
         for(i=; i<=n; i++)
             scanf("%s",map[i]+);
         Init_st_en();
         Init();
         ans = DFS();
         if(ans==-)
             printf("YouBadbad\n");
         else
             printf("%d\n",ans);
     }
     return ;
 }

 int DFS()
 {
     int i;
     int x,y,di,t;
     priority_queue<Point> q;
     Point cur,next;
     q.push(st);
     map[st.x][st.y] = '*';
     while(!q.empty())
     {
         cur = q.top();  q.pop();
         if(cur.x==en.x && cur.y==en.y)
             return cur.t;
         for(i=; i<; i++)
         {
             next.x = x = cur.x+dir[i][];
             next.y = y = cur.y+dir[i][];
             next.di= i;
             if(x< || x>n || y< || y>m || map[x][y]=='*') continue;
             di = cur.di;  //若从点cur到点next，到next的时候，方向一定是i
             t = cur.t;
             if(t!= && t%p==)
                 di = (di+-)%;
             if(di==i)  //可以1s从点cur到点next的条件
                 t++;
             else if(    ((t+)%p!= && ((di+)%==i || (di+-)%==i))  //可以2s从点cur到点next的条件
                     ||  ((t+)%p== && ((di+-)%==i || (di+-)%==i)))
                 t+=;
             else  //最多3s可以从点cur到点next
                 t+=;
             next.t = t;
             if(t<mark[x][y][i])
             {
                 q.push(next);
                 mark[x][y][i] = t;
             }
         }
     }
     return -;
 }