A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input
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 number of non-leaf nodes. Then M lines follow, each in the format:
ID K ID[1] ID[2] … ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID’s of its children. For the sake of simplicity, let us fix the root ID to be 01.
Output
For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.
The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output “0 1” in a line.
Sample Input
2 1
01 1 02
Sample Output
0 1
题目的意思就是通过一种特定格式的输入形成一棵树,要打印每一层的叶子结点的个数,可以用vector数组来存每个节点的child,每个vector代表一个节点,通过特定的输入来保存到vector中,在dfs中一旦发现某个vector为空说明该节点为叶子结点,那么就可以 book[当前floor]++,代表此floor的叶子节点个数加一。
#include<iostream>
#include <vector>
#include <algorithm>
using namespace std;
int m, n;
vector<int>v[100];
int book[100] = { 0 };
int height = 0, root;
void dfs(int index, int floor) {
if (v[index].size() == 0) {
book[floor]++;
if (height < floor)height = floor;
return;
}
for (int child : v[index]) {
dfs(child, floor + 1);
}
}
int main() {
cin >> n >> m;
for (int i = 0; i < m; i++) {
int k, node;
cin >> node >> k;
for (int m = 0; m < k; m++) {
int child;
cin >> child;
v[node].push_back(child);
}
}
dfs(1, 0);
cout << book[0];
for (int i = 1; i <= height; i++) {
cout <<" "<< book[i];
}
system("pause");
}
本文介绍了一种通过特定格式输入构建家族树的方法,并详细解释了如何使用C++编程语言来实现统计每层叶子节点数量的功能。
446

被折叠的 条评论
为什么被折叠?



