hdu1010

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 5S.X...X...XD....3 4 5S.X...X....D0 0 0

 

Sample Output

NOYES

 题意：有一张地图，地图里面有三种东西，起点‘S’，骨头'D',墙壁'X'，一只狗在里面走，问你这只狗能不能恰好在第t步到达。
思路：用深度优先去搜索，这里面要用到奇偶剪枝去剪枝。

现假设起点为(sx,sy)，终点为（ex,ey），给定t步恰好走到终点，

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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

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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）；

除此之外，就是简单的深度搜索题了。
代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;char map[10][10];int visited[10][10];int n,m,cnt,flag;int sx,sy,ex,ey;int dx[4]={1,-1,0,0},dy[4]={0,0,1,-1};void dfs(int x,int y,int count){    if(flag)        return;    if(x<0||x>=n||y<0||y>=m||map[x][y]=='X'||visited[x][y])        return;    if(x==ex&&y==ey){        if(count==cnt)//恰好到达            flag=1;        return;    }    int step=fabs(x-ex)+fabs(y-ey)+count;    if(step>cnt||(cnt-step)&1)//奇偶剪枝        return;    for(int i=0;i<4;i++){        visited[x][y]=1;        dfs(x+dx[i],y+dy[i],count+1);        visited[x][y]=0;//一定要记得还原    }}int main(){    while(~scanf("%d%d%d",&n,&m,&cnt)&&(n+m+cnt))    {        flag=0;        for(int i=0;i<n;i++){            scanf("%s",map[i]);            for(int j=0;j<m;j++){                if(map[i][j]=='S')                    sx=i,sy=j;                if(map[i][j]=='D')                    ex=i,ey=j;            }        }        memset(visited,0,sizeof(visited));        int step=fabs(sx-ex)+fabs(sy-ey);//最少步数        if(step>cnt||(cnt-step)&1)//奇偶剪枝            printf("NO\n");        else{            dfs(sx,sy,0);            if(flag)                printf("YES\n");            else                printf("NO\n");        }    }    return 0;}