Tempter of the Bone

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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 8
 
using namespace std;
 
bool flag,ans,visited[N][N];
//visited为标记是否可以访问的数组 
//ans为标记小狗是否能逃离，true-->yes  
//flag与ans作用一样，它是用来看看是否退出dfs函数 
int n,m,t,xe,ye;
//n,m,t分别为行数，列数，时间 
// (xe,xy)-->大门坐标 
char map[N][N]; //地图 
 
void dfs(int x,int y,int timen){
    if(flag) 
		return ;
    if(timen>t) 
    //时间超过给定时间 
		return ;
    if(x<0||x>n-1||y<0||y>m-1)
	//超过给定的范围 
		return ;
    if(timen==t&&map[x][y]=='D'){
    //时间恰好是给定的时间，且位置恰好在大门位置 
		flag=ans=true;
		return ;
	}
    int temp=t-timen-abs(xe-x)-abs(ye-y);
    /*
    *奇偶剪枝：
	*	是数据结构的搜索中，剪枝的一种特殊小技巧。
	*	现假设起点为(sx,sy)，终点为（ex,ey），给定t步恰好走到终点， 
	*		s 0 0 0 0
	*		| 0 0 0 0
	*		| 0 0 0 0
	*		| 0 0 0 0
	*		+ - - - e
	*	如图所示（"|"竖走，"-"横走，"+"转弯），易证abs(ex-sx)+abs(ey-sy)为此问题类中任意情况下，起点到终点的最短步数， 
	*	记做step，此处step1=8；
　　    *		s - - - 0
	*		0 - - + 0
	*		| + 0 0 0
	*		| 0 0 0 0
	*		+ - - - e
	*	如图，为一般情况下非最短路径的任意走法举例，step2=14；
	*	step2-step1=6，偏移路径为6，偶数（易证）；
	*	故，若t-[abs(ex-sx)+abs(ey-sy)]结果为非偶数（奇数），则无法在t步恰好到达；
	*	返回，false；
	*	反之亦反。
    */
	if(temp&1) //奇偶剪枝,位运算判断是否为奇数，比mod(求余数)更快.
	/*
	* 一个整数a, a & 1 这个表达式可以用来判断a的奇偶性。
	* 二进制的末位为0表示偶数，最末位为1表示奇数。
	* 使用a%2来判断奇偶性和a & 1是一样的作用，但是a & 1要快好多。
	*/
    	return ;
    	
    //向四个方向进行深搜	
	if(!visited[x-1][y]&&map[x-1][y]!='X'){
        visited[x-1][y]=true;
        dfs(x-1,y,timen+1);
        visited[x-1][y]=false;
    }
    if(!visited[x+1][y]&&map[x+1][y]!='X'){
        visited[x+1][y]=true;
        dfs(x+1,y,timen+1);
        visited[x+1][y]=false;
    }
    if(!visited[x][y-1]&&map[x][y-1]!='X'){
        visited[x][y-1]=true;
        dfs(x,y-1,timen+1);
        visited[x][y-1]=false;
    }
    if(!visited[x][y+1]&&map[x][y+1]!='X'){
        visited[x][y+1]=true;
        dfs(x,y+1,timen+1);
        visited[x][y+1]=false;
    }
}
/*
*		演示：input: 3 4 5
*			S . X .
*			. . X .
*			. . . D
*		S-->(0,0) D-->(3,2) cnt=2;
*		dfs(0,0,0) 
*		|         		|
*	    visit[0][1]=true
*	    dfs(0,1,1)    		dfs(1,0,1)
*		|
*	    visit[1][1]=true
*	    dfs(1,1,2)
*		|
*	    dfs(2,1,3)
*		|
*	    dfs(2,2,4)
*		|
*	    dfs(2,3,5)
*
*/
 
int main(){
    int xs,ys;//起点坐标(xs,ys) 
    while(scanf("%d%d%d",&n,&m,&t)!=EOF&&(n||m||t)){
    //当输入不是EOF和其中全部都不是0，程序运行 
        int cnt=0; 
        memset(visited,false,sizeof(visited)); //初始化 
        flag=ans=false;
        
        for(int i=0;i<n;i++){
        	scanf("%s",&map[i]);
            for(int j=0;j<m;j++){
                if(map[i][j]=='S'){
                //找出狗的初始位置，并将其位置标记为可以访问 
                    xs=i;ys=j;
                    visited[i][j]=true;
                }
                if(map[i][j]=='D'){
                //找出门的位置 
                    xe=i;ye=j;
                }
                if(map[i][j]=='X'){
                //找出墙的位置 
                    cnt++;//记录有多少堵墙 
                }
            }
        }
        if(n*m-cnt-1>=t) 
        //总格子数-墙的数量-狗的初始位置<给的时间数，能表明什么呢？ 
			dfs(xs,ys,0);
        
		if(ans) 
			printf("YES\n");
        else 
			printf("NO\n");
    }
    return 0;
}