题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1010 Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 131057    Accepted Submission(s): 35308 Problem Description The doggie fou…

http://acm.hdu.edu.cn/showproblem.php?pid=1010 这题就是问能不能在t时刻走到门口,不能用bfs的原因大概是可能不一定是最短路路径吧. 但是这题要过除了细心外,还需要强力的剪枝. 奇偶性剪枝:参考 http://www.cppblog.com/Geek/archive/2010/04/26/113615.html #include <iostream> #include <cstring> #include <cstdio>…

http://acm.hdu.edu.cn/showproblem.php?pid=1010   //题目链接 http://ycool.com/post/ymsvd2s//一个很好理解剪枝思想的博客 http://blog.csdn.net/chyshnu/article/details/6171758//一个很好举例的博客 Problem Description The doggie found a bone in an ancient maze, which fascinated him…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 82702    Accepted Submission(s): 22531 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

Tempter of the Bone Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 5   Accepted Submission(s) : 1 Problem Description The doggie found a bone in an ancient maze, which fascinated him a lot. Howe…

M - Tempter of the Bone Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u 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 d…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 92175    Accepted Submission(s): 25051 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1010 题目大意: 输入 n m t,生成 n*m 矩阵,矩阵元素由 ‘.’ 'S' 'D' 'X' 四类元素组成. S'代表是开始位置: 'D'表示结束位置:'.'表示可以走的路:'X'表示是墙. 问:从‘S’  能否在第 t 步 正好走到 'D'. 解题思路: 平常心态做dfs即可,稍微加个奇偶剪枝,第一次做没经验,做过一次下次就知道怎么做了.最后有代码注释解析. AC Code: #includ…

Tempter of the Bone Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up,…

题意:有一副二维地图'S'为起点,'D'为终点,'.'是可以行走的,'X'是不能行走的.问能否只走T步从S走到D? 题解:最容易想到的就是DFS暴力搜索,,但是会超时...=_=... 所以,,要有其他方法适当的剪枝:假设当前所在的位置为(x,y),终点D的位置为(ex,ey); 那么找下规律可以发现: 当 |x-ex|+|y-ey| 为奇数时,那么不管从(x,y)以何种方式走到(ex,ey)都是花费奇数步:当为偶数时同理.  这即是所谓的奇偶性剪枝.这样剪枝就可以复杂度就变为原来暴力DFS的开…