题目与翻译 1004 Counting Leaves 数树叶 (30分) A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. 一个家族的等级通常是由一个系谱树表示的。你的工作是统计那些没有孩子的家庭成员
每个输入文件包含一个测试用例。每个案例以包含0 <N <100(树中节点数)和M(<N)(非叶节点数)的行开头。 然后接下来的M行,每个格式为::ID K ID[1] ID[2]…ID [K]其中ID是一个表示给定非叶节点的两位数,K是其子节点的个数,后面是其子节点的一...
A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Specification: Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the num...
PAT (Advanced Level) Practice 1004 Counting Leaves (30 分) 非 DFS / BFS 解法 凌宸1642 题目描述: A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.. 译:一个家族的层次通常可以由一颗族谱树提现出来。你的任务就是找出那些...
voidcount_leaves(introot,intlayer=1){maxlayer=maxlayer>layer?maxlayer:layer;intleave_node=HierachyInfo[root].size();for(inti=0;i<HierachyInfo[root].size();i++)//遍历某一个根节点下的全部叶子节点{if(HierachyInfo[HierachyInfo[root][i]].size()>0)//如果某一个叶子节点下还有叶子节点{leave...
1004 Counting Leaves (30 分) 1004 Counting Leaves (30 分) 两种主流做法,第一种是将树用二叉树的方式存储(左二子右兄弟),最后斜着遍历。第二种直接将树存成本来的样子,之后做层次遍历,这里采用第二种方法。代码如下: ps:由于时间要求不高,可以将每一个树节点的指针都存到一个数组里,求出每一个节点的...
1004 Counting Leaves (30分)作者:CHEN, Yue 单位:浙江大学 代码长度限制:16 KB 时间限制:400 ms 内存限制:64 MB A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.Input Specification: Each input file contains one test case....
1004 Counting Leaves (30 分) A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Specification: Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a...
【PAT 甲级】1004 Counting Leaves (30分)(dfs水) 题目 给一颗节点数不超过 100 的树,求每一层的叶子节点的个数。 分析 建树之后记录每个节点有无孩子,之后 dfs 一下记录每个节点在第几层。将信息归总统计一下即可得出每层叶子节点数。 #include <bits/stdc++.h>...