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

 

又是一道搜索题目。使用深度优先搜索，注意要加入剪枝条件。不然非常easy造成超时，同一时候做搜索题要注意回溯。举个样例，若是求经典问题--油田的块数，我们不须要回溯，可是本题是求路径，我们须要用到回溯，不然会答案错误的，这是因为求油田块数的话，我们遍历了，不须要再次遍历，而求路径的话，若是求得的路径不通。须要将标记的路径还原，回溯非常重要！

！

。。

本题代码例如以下：

#include <iostream>

#include <string>

#include <cmath>

#include <cstring>

using namespace std;

int N,M,T,escape,wall;

int starti,startj,endi,endj;

char map[101][101];

int v[101][101];

int dir[4][2]={1,0,-1,0,0,1,0,-1};

void dfs(int x,int y,int time)

{

    if(abs(endi-x)+abs(endj-y)>T-time)return;

    if((abs(endi-x)+abs(endj-y))%2!=(T-time)%2)return;

    if(x==endi&&y==endj&&time==T)escape=1;

    if(escape==1)return;

    for(int i=0;i<=3;i++)

    {

        int xx=x+dir[i][0];

        int yy=y+dir[i][1];

        if(xx>=1&&xx<=N&&yy>=1&&yy<=M&&map[xx][yy]!='X'&&v[xx][yy]==0)

        {

            v[xx][yy]=1;

            dfs(xx,yy,time+1);

            v[xx][yy]=0;   //回溯

        }

    }

    return;

}

int main()

{

    while(cin>>N>>M>>T,N||M||T)

    {

        wall=0;

        memset(v,0,sizeof(v));

        for(int i=1;i<=N;i++)

            for(int j=1;j<=M;j++)

            {

                cin>>map[i][j];

                if(map[i][j]=='S')

                {

                    starti=i;

                    startj=j;

                }

                else if(map[i][j]=='X')

                {

                    wall++;

                }

                else if(map[i][j]=='D')

                {

                    endi=i;

                    endj=j;

                }

            }

        v[starti][startj]=1;

        escape=0;

        dfs(starti,startj,0);

        if(N*M-wall<T){cout<<"NO"<<endl;continue;}

        if(escape==1)cout<<"YES"<<endl;

        else cout<<"NO"<<endl;

    }

    return 0;

}