HDU 1010 Tempter of the Bone(深搜：奇偶剪枝)

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

//需要求的是能否在规定的那个时间刻刚好到达，而不是在规定时间内能否到达：（该代码不小心会超时的~~~~）
#include<stdio.h>
#include<string.h>
#include<math.h>
#define N 30
int m, n, T, flag, c, d, mark[N][N];
char Map[N][N];
int dir[4][2] = { {1,0}, {-1,0}, {0,1}, {0,-1} };
int Dis(int x, int y)
{
    int a, b;
    a = fabs(x-c);
    b = fabs(y-d);
    return a+b; //求出该点到终点的最短距离
}
int Judge(int nx, int ny) //判断该点是否能走
{
    if (nx >= 0 && nx < m && ny >= 0 && ny < n && Map[nx][ny] != 'X' && mark[nx][ny] == 0)
        return 1;
    return 0;
}
void DFS(int x, int y, int s)
{
    int dis, nx, ny, i;
    if (s == T && x == c && y == d)
    {
        flag = 1;
        return ;
    } //能够到达则停止搜索
    if(flag) return ;
    dis = Dis(x, y);
    if (dis > T-s || (dis+T-s)%2 != 0) return ; //要是最短时间比还能走的时间大，就不能继续，当要走的步数和还剩下的步数奇偶不同时也不能继续（奇偶剪枝）
    for (i = 0; i < 4; i++)
    {
        nx = x + dir[i][0];
        ny = y + dir[i][1];
        if (Judge(nx, ny))
        {
            mark[nx][ny] = 1; //用mark数组标记是否到达过，更节约时间
            DFS(nx, ny, s+1);
            mark[nx][ny] = 0;
        }
    }
}
int main ()
{
    int i, j, a, b;
    while (scanf("%d %d %d", &m, &n, &T), m+n+T)
    {
        memset(mark, 0, sizeof(mark));
        flag = 0;
        for (i = 0; i < m; i++)
            scanf("%s", Map[i]);
        for (i = 0; i < m; i++)
        {
            for (j = 0; j < n; j++)
            {
                if(Map[i][j] == 'S')
                {
                    a = i;
                    b = j;
                }
                if (Map[i][j] == 'D')
                {
                    c = i;
                    d = j;
                }     
            }
        }
        mark[a][b] = 1;
        DFS(a, b, 0);
        if (flag == 1) printf("YES
");
        else printf("NO
");
    }
    return 0;
}

下面有个不会超时的：（其实两个代码意思差不多，都运用了奇偶剪枝，但是第二个更剪枝。。。）

#include<stdio.h>
#include<string.h>
#define N 30
int m, n, T, flag, c, d, mark[N][N];
char Map[N][N];
int dir[4][2] = { {1,0}, {-1,0}, {0,1}, {0,-1} };
int Dis(int a);
void DFS(int x, int y, int s);
int main ()
{
    int i, j, a, b, sum;
    while (scanf("%d %d %d", &m, &n, &T), m+n+T)
    {
        sum = 0;
        memset(mark, 0, sizeof(mark));
        flag = 0;
        for (i = 0; i < m; i++)
            scanf("%s", Map[i]);
        for (i = 0; i < m; i++)
        {
            for (j = 0; j < n; j++)
            {
                if(Map[i][j] == 'S')
                {
                    a = i;
                    b = j;
                }
                if (Map[i][j] == 'D')
                {
                    c = i;
                    d = j;
                }
                if(Map[i][j] == 'X')
                {
                    sum++;
                }
            }
        }
        if(n*m-sum-1 < T) printf("NO
");
        else
        {
            mark[a][b] = 1;
            DFS(a, b, 0);
            if (flag == 1) printf("YES
"); 
            else printf("NO
");
        }
    }
    return 0;
}
int Dis(int a)
{
    return a >= 0 ? a : -a;
}
void DFS(int x, int y, int s)
{
    int dis, nx, ny, i;
    if (s == T && x == c && y == d)
    {
        flag = 1;
        return ;
    }
    if(flag) return ;
    dis = Dis(x-c) + Dis(y-d);
    if (dis > T-s || (dis+T-s)%2 != 0) return ;
    for (i = 0; i < 4; i++)
    {
        nx = x + dir[i][0];
        ny = y + dir[i][1];
        if (nx >= 0 && nx < m && ny >= 0 && ny < n && Map[nx][ny] != 'X' && mark[nx][ny] == 0)
        {
            mark[nx][ny] = 1;
            DFS(nx, ny, s+1);
            mark[nx][ny] = 0;
        }
    }
}