题目传送门 题意:给出线性排列的树,第一个数字是根节点,后面的数如果当前点小或相等往左走,否则往右走,查询一些点走的路径 分析:题意略晦涩,其实就是排序二叉树!1<<1000 普通数组开不下,用指针省内存.将树倒过来好理解些 E---------------------------------------W 2 / \ 1 4 / 3 6 / 5 / 4 / 3 / 2 / 1 E---------------------------------------W 收获:排序二叉树插入和查询 代码…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5444 Elven Postman Description Elves are very peculiar creatures. As we all know, they can live for a very long time and their magical prowess are not something to be taken lightly. Also, they live on tr…

题目链接:pid=5444http://">http://acm.hdu.edu.cn/showproblem.php?pid=5444 Elven Postman Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1206    Accepted Submission(s): 681 Problem Description E…

Time Limit: 1500/1000 MS (Java/Others)   Memory Limit: 131072/131072 K (Java/Others) Problem Description Elves are very peculiar creatures. As we all know, they can live for a very long time and their magical prowess are not something to be taken lig…

Problem Description Elves are very peculiar creatures. As we all know, they can live for a very long time and their magical prowess are not something to be taken lightly. Also, they live on trees. However, there is something about them you may not kn…

Elven Postman Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 591    Accepted Submission(s): 329 Problem Description Elves are very peculiar creatures. As we all know, they can live for a very…

Elven Postman Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5444 Description Elves are very peculiar creatures. As we all know, they can live for a very long time and their magical prowess are not something to…

Elven Postman Elves are very peculiar creatures. As we all know, they can live for a very long time and their magical prowess are not something to be taken lightly. Also, they live on trees. However, there is something about them you may not know. Al…

题意:给出一颗二叉树的先序遍历,默认的中序遍历是1..2.……n.给出q个询问,询问从根节点出发到某个点的路径. 析:本来以为是要建树的,一想,原来不用,其实它给的数是按顺序给的,只要搜结点就行,从根开始搜,如果要到的结点比根结点大,那么一定是向W走, 然后去第一个结点,然后接着判定,一直走,如果找到结束就好.水题.当时想的有点复杂. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #includ…

HDU 5444 题意:给你一棵树的先序遍历,中序遍历默认是1...n,然后q个查询,问根节点到该点的路径(题意挺难懂,还是我太傻逼) 思路:这他妈又是个大水题,可是我还是太傻逼.1000个点的树,居然用标准二叉树结构来存点,,,卧槽想些什么东西.可以用一维数组,left,right直接指向位置就行了,我这里用的是链表.对于读入的序列,生成一个二叉排序树,同时记录一下路径就可以了,后面居然忘了清空path又wa了一发. #include<iostream> #include<cstdio…