HDOJ1010 (DFS+奇偶剪枝)

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hdoj--1010<dfs+奇偶剪枝>

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1010 题目描述:在n*m的矩阵中,有一起点和终点,中间有墙,给出起点终点和墙,并给出步数,在该步数情况下走到终点,走过的点不能再走: 题目要点:dfs+奇偶剪枝:输入: 本题用dfs可以做出结果,但是会超时,需要用到就剪枝,来减去大部分的可能: 奇偶剪枝: 方格中起点(tx,ty)和终点(dx, dy)最小步骤是minstep=abs(tx-dx)+abs(ty-dy); 给定步数t,从起点走到终点…

HDOJ1010 (DFS+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): 107043 Accepted Submission(s): 29107 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

HDU 1010 Tempter of the Bone(DFS+奇偶剪枝)

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…

hdu.1010.Tempter of the Bone(dfs+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): 82702 Accepted Submission(s): 22531 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

hdu 1010:Tempter of the Bone（DFS + 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): 58766 Accepted Submission(s): 15983 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

杭电1010（dfs + 奇偶剪枝）

题目: 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 ou…