zoj 2110 Tempter of the Bone_不见不散的结局是曲终人散

Tempter of the Bone

Time Limit: 2 Seconds Memory Limit: 65536 KB

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：S代表出发位置，D代表终点位置，X代表不能走，'.'可以走，输入第一行的三个数字N,M，T。气质N代表行数，M代表列数，T代表需要的时间，求是否能刚好T秒从S走到D；

思路：原先也是直接DFS，但是一直超时，最后在网上看，才知道要要奇偶性减枝，关于这个原理，自己去模拟一下就知道了；

代码如下：

#include < stdio.h >
#include < stdlib.h >
#include < stdbool.h >
bool flag;
int n,m,t,i,j,x1,y1,x2,y2,num;
int dir[4][2]= {{0,-1},{0,1},{1,0},{-1,0}}; //设置四个方向，分别是左、右、下、上
char str[8][8];
void BFS(int k,int l,int tc)
{
    int i;
    if(tc==t && k==x2 &&l==y2)//到终点时，判断时间步数是否相等
        flag=1;
    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 dx=k+dir[i][0];
        int dy=l+dir[i][1];
        if(dx >= 0 && dx < n && dy >= 0 && dy < m && str[dx][dy] != 'X')
        {
            str[dx][dy]='X';
            BFS(dx,dy,tc+1);    //开始深度优先搜索
            str[dx][dy]='.';    //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;//用来计算坐标上面可以走最多的步数；
        getchar();
        flag=0;
        for(i=0; i < n;i++)
        {
            for(j=0; j < m;j++)
            {
                scanf("%c",&str[i][j]);
                if(str[i][j]=='S')//记录初始位置；
                {
                    x1=i;
                    y1=j;
                }
                else if(str[i][j]=='D')//记录终点位置；
                {
                    x2=i;
                    y2=j;