HDU1010:Tempter of the Bone(dfs+剪枝)-白红宇

HDU1010:Tempter of the Bone(dfs+剪枝)

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发布时间：2019-06-27

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//一个很好理解剪枝思想的博客

//一个很好举例的博客

Problem Description

The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.

The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:

'X': a block of wall, which the doggie cannot enter;

'S': the start point of the doggie;

'D': the Door; or

'.': an empty block.

The input is terminated with three 0's. This test case is not to be processed.

Output

For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.

Sample Input

4 4 5

S.X.

..X.

..XD

....

3 4 5

S.X.

..X.

...D

0 0 0

Sample Output

YES

例题：

题目意思是讲有一只狗要吃骨头，结果进入了一个迷宫陷阱，迷宫里每走过一个地板费时一秒，该地板就会在下一秒塌陷，所以你不能在该地板上逗留。迷宫里面有一个门，只能在特定的某一秒才能打开，让狗逃出去。现在题目告诉你迷宫的大小和门打开的时间，问你狗可不可以逃出去，可以就输出YES,否则NO。

搜索时要用到的剪枝：

1.如果当前时间即步数(step) >= T 而且还没有找到D点，则剪掉。

2.设当前位置(x, y)到D点(dx, dy)的最短距离为s，到达当前位置(x, y)已经花费时间(步数)step，那么，如果题目要求的时间T - step < s，则剪掉。

3. 对于当前位置(x, y)，如果，(T-step-s)是奇数，则剪掉(奇偶剪枝)。

4.如果地图中，可走的点的数目(xnum) < 要求的时间T，则剪掉(路径剪枝)。

题目解析：

通过做这题算是懂剪枝的思想了，要学奇偶剪枝首先要看懂那个01矩阵（很好理解），之后就没什么问题了，

剪完枝后大约100ms就过了，怎么说呢，还是了解思想比较重要。

#include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            using namespace std;int n,m,t,tx,ty,flag;char map[8][8];int v[8][8];int jx[]= {
     1,-1,0,0};int jy[]= {
     0,0,1,-1};int Distance ( int x, int y ){    return abs(x - tx )+abs( y - ty ); // 当前点(x,y)到终点(tx,ty)的最短距离}void dfs(int x,int y,int ans){    if(ans>t) return ;    if(x==tx&&y==ty&&ans==t)    {        flag=1;        return ;    }    int dis=t-ans-Distance(x,y);    if(dis<0||dis%2) return ;// 剩余步数小于最短距离或者满足奇偶剪枝条件      for(int i=0; i<4; i++)    {        int xx=x+jx[i];        int yy=y+jy[i];        if(xx>=0&&xx
            
             =0&&yy
             
              t) { dfs(xx,yy,0);// 可通行的点必须大于要求的步数，路径剪枝。 } if(flag==1) printf("YES\n"); else printf("NO\n"); } return 0;}