hdu——1010Tempter of the Bone

Tempter of the Bone
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 102253    Accepted Submission(s): 27703






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




 




Author


ZHANG, Zheng


 




Source


 ZJCPC2004  


 




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JGShining

一道简单的dp模板题  算是自己可以编的唯一一道dp  代码是杭电discuss是一位大神的  解释比较详细就copy了一下  

有一个这样的代码是用来减枝的

if ((abs(di-si)+abs(dj-sj))%2!=(T-t)%2) return ;

注意，这道题目是要恰好t时间到达，并不是在t时间内到达......



第一个剪枝我们可以想到，当剩下的步数大于剩下的时间的时候，狗是不能走到的；
 
接下来我们来第二个剪枝：
我们把map的奇偶性以01编号：
0 1 0 1 0 1 
1 0 1 0 1 0 
0 1 0 1 0 1 
1 0 1 0 1 0 
0 1 0 1 0 1 
我们发现从0走一步一定走到1，从1走一步一定走到0。
也就是说，如果当前的狗所在的坐标与D的坐标奇偶性不一样，那么狗需要走奇数步。
同理，如果狗所在坐标与D的坐标奇偶性一样，那么狗需要走偶数步数。
 
也就是说，狗的坐标x、y和对2取余是它的奇偶性，Dxy和对2取余是D的奇偶性。
两个奇偶性一加再对2取余，拿这个余数去与剩下时间对2取余的余数作比较即可。

//http://blog.sina.com.cn/s/blog_a0d75e850100ydgs.html//不要骂我 哈哈哈哈// 求大神解释剪枝 。。。。。#include<stdio.h>#include<math.h>#include<algorithm>int dx[]={-1,0,1,0};int dy[]={0,1,0,-1};int di,dj;int si,sj;int is;char map[10][10];int n,m,T,word;void dfs(int di,int dj,int t)//在这我们传了三个量 一个是狗狗位置 x，y 还有一个就是该死的出题者给的步数{int i;if(di<1||dj<1||di>n||dj>m||t>T||is)//这里不知道各位有没有看懂 就是杂七杂八的判断了一些不可能的情况        return; // 比如有狗狗 眼睛不好走到地图外的情况或者超时的情况if(si==di && sj==dj && t==T) //在这里狗狗就要笑了 小样还想关我  狗狗艰难的爬过了九九八十一难并且在规定的时间内找到了出口我们就给它一个胜利的标记    {        is=1;        return ;    }if ((abs(di-si)+abs(dj-sj))%2!=(T-t)%2) return ;//这里是剪枝 用来优化的 去掉一些不可能的情况不用在进行搜索的情况 我也不懂求大神教for(i=0;i<4;i++)//这边就控制狗狗的四个方位递归思想回溯{if(map[di+dx[i]][dj+dy[i]]!='X')//判断狗狗周围的路是不是墙 是墙那狗狗也不是笨狗 当然还是会绕路走的{map[di+dx[i]][dj+dy[i]]='X';//如果这条路不是墙 嘿嘿 狗狗就昂头挺胸甩着尾巴 大步向前走 妹妹你推车哥哥大步向前走；dfs(di+dx[i],dj+dy[i],t+1);//当然狗狗每走一步我们就要在这个点上向之前一样再进行四个方向的判断map[di+dx[i]][dj+dy[i]]='.';}}return;}int main(){int i,j;while(~scanf("%d%d%d",&n,&m,&T)&&n!=0){word=0;getchar();for(i=1;i<=n;i++)//这里是输入不用多说了{for(j=1;j<=m;j++){scanf("%c",&map[i][j]);if(map[i][j]=='S')//我们把 狗狗的初始位置记录下来{di=i;dj=j;}if(map[i][j]=='D')//出口的初始位置记录{si=i;sj=j;}if(map[i][j]=='X')//墙的数量也记录下来{word++;}}getchar();}is=0;if(n*m-word<T)//地图上 所有的点 减去 墙的数量 小于 给定的步数那么肯定可以过的is=0;//因为 如果所有的点都在规定步数中走完还走不到那个点 要么那个点周围的墙都是封死的这狗狗会在孤独中死去因为没法进门了！！else// 要么就是没有这个门这个更狠了 这狗狗在这图上走好久好久才发现该死的出题者没有设门！！ 这个两个都是错误的图题目不会出{map[di][dj]='X';//我们把迈出的第一步变成墙封死这样就防止这条狗走回路 他知道是墙就不会给墙撞死 dfs(di,dj,0);//这里我们到上面的函数去}if(is==1)printf("YES\n"); //GOODelseprintf("NO\n");//DARK!!}}