HDOJ1010DFS（加剪枝）

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

NO

YES

http://acm.hdu.edu.cn/showproblem.php?pid=1010

题意就是在规定步数从S点能不能到D点 （注意一步不多一步不少）

深搜广搜都可以 下面是深搜

用到奇偶剪枝 最短距离也就是曼哈顿距离 必须 和输入步数同 奇偶 才能到 因为 你绕x或Y 多走一步就必须绕回来一步 

所以只要回到那个终点 所有的远路必须和最短路 奇偶一样

#include
#include
#include
int dir[4][2]= {1,0,0,1,-1,0,0,-1};
int vis[8][8];
int gx,ggx,gy,ggy,t,flag,n,m;
char ap[10][10];
void dfs(int x,int y,int k)
{
    int i;
    if(k>t) return ;
    if(x==ggx&&y==ggy)
    {
        if(k==t)
        {
            printf("YES\n");
            flag=1;
        }
        return ;
    }
    for(i=0; i<4; i++)
    {
        int xx=x+dir[i][0];
        int yy=y+dir[i][1];
        if(ap[xx][yy]!='X'&&xx>=0&&xx=0&&yy

还有个 剪枝 记录墙的数量如果n*m-墙数<=步数 也不可能