HDOJ 1010 Tempter of the Bone

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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

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4 4 5 
S.X. 
..X. 
..XD 
.... 
3 4 5 
S.X. 
..X. 
...D 
0 0 0

Sample Output

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NO 
YES

奇偶性剪枝

我们可以把map看成是下图中的样子：

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0 1 0 1 0 1
1 0 1 0 1 0
0 1 0 1 0 1
1 0 1 0 1 0
0 1 0 1 0 1

从0走一步必然走到1，从1走一步必然走到0

即：

  • 0->1或1->0 必然为奇数步
  • 0->0或 1->1必然为偶数步

所以： 当遇到从0走向0而要求时间为奇数的，或者0走向1要求时间为偶数的情况，可直接判断NO

此外

由距离公式以及上述推断，当前位置与所剩时间的奇偶性也要相同。

即剩余的时间t的奇偶性必定与abs(xi - x0) + abs(yi - y0)的奇偶性相同。不然就无法到达

最后就是DFS加上奇偶性剪枝

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#include<stdio.h>
#include<stdlib.h>
bool flag;
int n,m,t,i,j,x1,y1,x2,y2,num;
int dx[4]={0,0,-1,1};
int dy[4]={-1,1,0,0};  //设置四个方向，分别是左、右、下、上
char str[8][8];
void dfs(int k,int l,int tc)
{
    int i;
    if(tc==t && k==x2 &&l==y2)
       flag=true;
    if(flag) 
	   return;  
    if((abs(x2-k)+abs(y2-l))%2!=(t-tc)%2) 
	   return;   //奇偶性剪枝的判断条件
    for(i=0;i<4;i++)
    {
        int xx=k+dx[i];
        int yy=l+dy[i];
        if(xx>=0 && xx<n && yy>=0 && yy<m && str[xx][yy]!='X')
        {
            str[xx][yy]='X';
            dfs(xx,yy,tc+1);
            str[xx][yy]='.';    //此方向如果不行的话就回溯
        }
    }
}
int main()
{
    while (scanf("%d%d%d",&n,&m,&t))
    {
        if(n==0 && m==0 && t==0)
            break;
        num=0;
        flag=false;
        for(i=0;i<n;i++)
        {
        	scanf("%s",str[i]);
            for(j=0;j<m;j++)
            {
                if(str[i][j]=='S')
                {
                    x1=i;y1=j;
                }
                else if(str[i][j]=='D')
                {
                    x2=i;y2=j;
                    num++;
                }
                else if(str[i][j]=='.')
                {
                    num++;
                }
            }
        }
        str[x1][y1]='X';
        if(num>=t)
          dfs(x1,y1,0);
        if(flag) 
            printf("YES\n");
        else 
            printf("NO\n");
    }
    return 0;
}
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