Tempter of the Bone

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

题意：小狗找出口，给你一个n*m的地图和小狗，门的位置，让你小狗能否刚好在t时到达门口；

解题思路：刚开始想的是用广搜求出最小时间，如果最小时间

另外还可以加一个剪枝条件，当然了不加这个也能AC，(n*m-s<=t)，t过大的时候肯定不能按时到达了；

感悟：深搜好些但是，剪枝不好写啊，不剪枝就会超时，好烦的！

代码：

#include

#include

#include

#include

#include

#define maxn 205

using namespace std;

int visit[maxn][maxn];//这个是记录步数的不是记录走没走的不能用布尔型

char mapn[maxn][maxn];

int n,m,t,direction[4][2]={{1,0},{-1,0},{0,1},{0,-1}};

int sx,sy,dx,dy;

bool flag;

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

{

   
if(x<1||x>n||y<1||y>m)//边界

       
return;

   
if(flag==1)

       
return;

   
if(x==dx&&y==dy&&time==t)//找到D了

    {

       
if(time=t)

           
flag=1;

       
return;

    }

    int
temp=(t-time)-abs(x-dx)-abs(y-dy);//奇偶性剪枝

   

   
if(temp<0||temp&1)  return;

   

   

    for(int
i=0;i<4;i++)

    {

       
int x1=x+direction[i][0];

       
int y1=y+direction[i][1];

       
if(mapn[x1][y1]!='X')

       
{

           
mapn[x1][y1]='X';

           
dfs(x1,y1,time+1);

           
mapn[x1][y1]='.';//不满足条件的话就返回

       
}

    }

   
return;

}

int main()

{

    int
s=0;

   
//freopen("in.txt", "r", stdin);

   
while(~scanf("%d%d%d\n",&n,&m,&t)&&n&&m&&t)

{

       
flag=s=0;

       
for(int i=1;i<=n;i++)

       
{

           
for(int j=1;j<=m;j++)

           
{

               
scanf("%c",&mapn[i][j]);

               
if(mapn[i][j]=='S')

               
{

                   
sx=i;

                   
sy=j;//狗的位置

               
}

               
if(mapn[i][j]=='D')

               
{

                   
dx=i;

                   
dy=j;//门的位置

               
}

               
if(mapn[i][j]=='X')

                   
s++;

           
}

           
scanf("\n");

       
}

       
mapn[sx][sy]='X';

       
memset(visit,0,sizeof(visit));

       
//for(int i=1;i<=n;i++)

       
//{

       
//    for(int
j=1;j<=m;j++)

       
//    {

       
//       
printf("%c",mapn[i][j]);

       
//    }

       
//   
printf("\n");

       
//}

       
if (n*m-s<=t)//提前判断t过大的情况避免再去搜

       
{

           
printf("NO\n");

           
continue;

       
}

       
dfs(sx,sy,0);

       
//printf("最少走 %d 秒 有 %d 秒\n",ans,t);

       
if(flag)

           
printf("YES\n");

       
else

           
printf("NO\n");

    }

}