TZOJ 1221 Tempter of the Bone(回溯+剪枝)

【TZOJ 1221 Tempter of the Bone(回溯+剪枝)】的更多相关文章

TZOJ 1221 Tempter of the Bone(回溯+剪枝)

描述 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…

HDU1010 Tempter of the Bone(回溯 + 剪枝)

本文链接:http://i.cnblogs.com/EditPosts.aspx?postid=5398734 题意: 输入一个 N * M的迷宫,这个迷宫里'S'代表小狗的位置,'X'代表陷阱,‘D’代表门,‘.’代表可行走的地方,小狗每次可以选择往周围的四个方向行走,问这个小狗能否正好T步找到门. 思路: 利用回溯 + 剪枝,这道题剪枝特别重要. 剪枝一: 可以把图看成这样: 1 0 1 0 10 1 0 1 01 0 1 0 10 1 0 1 01 0 1 0 1 则假设从点 a(i +…

HDU1010:Tempter of the Bone(dfs+剪枝)

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…

hdu 1010 Tempter of the Bone 奇偶剪枝

如果所给的时间(步数) t 小于最短步数path,那么一定走不到. 若满足t>path.但是如果能在恰好 t 步的时候,走到出口处.那么(t-path)必须是二的倍数. 关于第二种方案的解释: 这种方案学名为“奇偶剪枝”.我们已知了最短的步数就是直角三角形的两条直角边,实际上的路径却不一定非要沿着这两条边走的.仔细看看只要是移动方向一直是右.下,那么走到的时候总步数也一定是path的.然而由于墙的存在或许我们不可能一直右.下的走下去.为了避开墙,我们可能会向左走,向上走等等.但为了到达目的地…

Tempter of the Bone dfs+剪枝

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…

B - Tempter of the Bone(DFS+剪枝)

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…