hdu 1010 Tempter of the Bone(深搜+奇偶剪枝)

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 615 Accepted Submission(s): 219

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

分析：
（1）题意很清楚，就是在t时间正好达到出口，使用的是深搜+剪枝
（2）关于深搜不再赘述，这里的深搜并不麻烦，麻烦的是剪枝
（3）一开始我也一直的超时，上网查了别人的解释才知道这道题需要剪枝，一个以前没见过的剪枝。。。。
奇偶剪枝（转自百度百科）：

是数据结构的搜索中，剪枝的一种特殊小技巧。 　　
现假设起点为(sx,sy)，终点为（ex,ey），给定t步恰好走到终点， 　　

s
|
|
|
+	—	—	—	e

如图所示（“|”竖走，“—”横走，“+”转弯），易证abs(ex-sx)+abs(ey-sy)为此问题类中任意情况下，起点到终点的最短步数，记做step，此处step1=8； 　　

s	—	—	—
	—	—	+
|	+
|
+	—	—	—	e

如图，为一般情况下非最短路径的任意走法举例，step2=14； 　　
step2-step1=6，偏移路径为6，偶数（易证）； 　　
故，若t-[abs(ex-sx)+abs(ey-sy)]结果为非偶数（奇数），则无法在t步恰好到达； 　　
返回，false； 　　
反之亦反。
（4）应用到这道题时，t-abs(r-end_x)-abs(c-end_y)应该与step的奇偶性相同。。所以t-step-abs(r-end_x)-abs(c-end_y)应该为偶数。。。。（易证，不多说了）
当然小于0时更不能到了，因为比最短的都小肯定不符合。。。。。

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;

//迷宫
char maze[10][10];
//标记此处是否走过，0为未走过，1为走过
int mark [10][10];

//四个方向 上 下 左 右
int direct[4][2] = {-1,0,1,0,0,-1,0,1};

//迷宫行列，和开门时间
int row,column,t;
//开始位置，和门的位置
int begin_x,begin_y,end_x,end_y;

//r,c 表示搜索的位置,step表示走的步子也等于时间
bool dfs(int r,int c,int step)
{
    int x,y;
    //成功了，成功逃脱
    if(r==end_x&&c==end_y&&step==t)
    return true;

    //剪枝，时间已经到了，逃脱失败
    if(step>=t)return false;
    //奇偶剪枝...........没有这个剪枝，超时。。。呜呜。。
    int temp=t-step-abs(r-end_x)-abs(c-end_y);
    if (temp<0 || temp%2!=0)return false;

    for(int i = 0; i < 4; i++)
    {
        x = r + direct[i][0];
        y = c + direct[i][1];
         //出界了
        if(x<0||x>=row||y<0||y>=column)
        continue;
        //此路不通
        if(maze[x][y]=='X')continue;
        //该位置已经走过了。。
        if(mark[x][y]==1)continue;
        mark[x][y] = 1;
        if(dfs(x,y,step+1))return true;
        mark[x][y] = 0;
    }
    return false;
}


int main()
{
    int wall;
    while(cin>>row>>column>>t&&row!=0&&column!=0&&t!=0)
    {
        wall = 0;
        //初始化，标记为未走过
        for(int i = 0; i < 10; i++)
        for(int j = 0; j < 10; j++)
        mark[i][j] = 0;
        for(int i = 0; i < row; i++)
        {
            scanf("%s",maze+i);
            for(int j = 0; j < column;j++)
            if(maze[i][j]=='S')
            {
                begin_x = i;
                begin_y = j;
            }else if(maze[i][j]=='D'){
                end_x = i;
                end_y = j;
            }else if(maze[i][j]=='X')wall++;
        }
        //剪枝一
        if(row*column-wall<=t)
        {
            printf("NO\n");
            continue;
        }
        mark[begin_x][begin_y]=1;
        if(dfs(begin_x,begin_y,0))
        printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}