HDU 1010 Tempter of the Bone （dfs+奇偶减枝）

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 106608    Accepted Submission(s): 28989

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
  

 

Author

ZHANG, Zheng

 

Source

ZJCPC2004

 

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注意：题上要求T分钟到达而不是T分钟内到达

奇偶剪枝：根据题目，doggie必须在第t秒到达门口。也就是需要走t-1步。设doggie开始的位置为（sx,sy）,目标位置为（ex,ey）.如果abs(ex-x)+abs(ey-y)为偶数，则abs(ex-x)和abs(ey-y)奇偶性相同，所以需要走偶数步； 

当abs(ex-x)+abs(ey-y)为奇数时，则abs(ex-x)和abs(ey-y)奇偶性不同，到达目标位置就需要走奇数步。先判断奇偶性再搜索可以节省很多时间，不然的话容易超时。t-sum为到达目标位置还需要多少步。因为题目要求doggie必须在第t秒到达门的位置，所以（t-step）和abs(ex-x)+abs(ey-y)的奇偶性必然相同。因此temp=（t-step）-abs(ex-x)+abs(ey-y)必然为偶数。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char map[10][10];
int vis[10][10];
int n,m,t,sx,sy,dx,dy,flag;
int dix[]={0,0,-1,1};
int diy[]={-1,1,0,0};
void dfs(int x,int y,int num)
{
	if(flag==1)
	return ;
	if(x==dx&&y==dy&&num==t)
	{
		flag=1;
		return ;
	}
	int minn=abs(x-dx)+abs(y-dy);
	if(minn>t-num||(minn+t-num)%2!=0)
	return ;
	for(int i=0;i<4;i++)
	{
		int a=x+dix[i];
		int b=y+diy[i];
		if(a>=0&&b>=0&&a<n&&b<m&&!vis[a][b]&&map[a][b]!='X')
		{
			 vis[a][b]=1;
		     dfs(a,b,num+1);
		     vis[a][b]=0;
		}
	}
}
int main()
{
	int i,j,k;
	while(scanf("%d%d%d",&n,&m,&t),n+m+t)
	{
		k=0;
	
		for(i=0;i<n;i++)
		{
			scanf("%s",map[i]);
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='S')
				{
					sx=i;
					sy=j;
				}
				if(map[i][j]=='D')
				{
					dx=i;
					dy=j;
				}
				if(map[i][j]=='X')
				{
					k++;
				}
			}
		}
		if(n*m-k-1<t)
		{
			printf("NO\n");
			continue;
		}
		flag=0;
		memset(vis,0,sizeof(vis));
		vis[sx][sy]=1;
		dfs(sx,sy,0);
		if(flag)
		printf("YES\n");
		else
		printf("NO\n");
	}
	return 0;
}