zoj 2110 Tempter of the Bone (dfs)

Tempter of the Bone

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

 

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Author: ZHANG, Zheng
Source: Zhejiang Provincial Programming Contest 2004

//C++    20    172    姜伯约
/*

    题意：
        从S走向D，问能否恰好在第t个时刻走到 
    
    DFS：
        比较经典的一道深搜，奇偶剪枝是比较亮的一点，亲可以自己手动画图
    比较一下，会获益不少，其他和普通的dfs差不多。 

*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h> 
int n,m,t;
char g[15][15];
int mov[4][2]={0,1,1,0,0,-1,-1,0};
int flag,ex,ey;
void dfs(int x,int y,int cnt)
{
    if(cnt>t || flag) return;
    if(x==ex && y==ey && cnt==t) flag=1;
    int temp=(t-cnt)-abs(x-ex)-abs(y-ey);
    if(temp<0 || (temp&1))return; //奇偶剪枝 
    for(int i=0;i<4;i++){
        int tx=x+mov[i][0];
        int ty=y+mov[i][1];
        if(tx>=0 && tx<n && ty>=0 && ty<m && g[tx][ty]!='X'){
            g[tx][ty]='X';
            dfs(tx,ty,cnt+1);
            g[tx][ty]='.';
        }
    }
}
int main(void)
{
    while(scanf("%d%d%d",&n,&m,&t)!=EOF&&(n+m+t))
    {
        int x,y;
        int cnt=0;
        for(int i=0;i<n;i++){
            scanf("%s",g[i]);
            for(int j=0;j<m;j++){
                if(g[i][j]=='S') 
                    x=i,y=j;
                else if(g[i][j]=='.') cnt++;
                else if(g[i][j]=='D') ex=i,ey=j;
            }
        } 
        if(cnt+1<t){
            puts("NO");continue;
        }
        flag=0;
        g[x][y]='X';
        dfs(x,y,0);
        if(flag) puts("YES");
        else puts("NO");
    }
    return 0;
}