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 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.
The input ends with N being 0. That case must NOT be processed.
Output Specification:
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
这一题求树叶子结点的个数
解释一下样例输入和输出
通过读题不难看出，这棵树只有两个节点1和2，根节点是1（题目有说），它的子节点是2，所以这棵树叶子结点只有一个就是2，根节点在第一层，它的叶子结点在第二层所以输出时
输出0和1（第一层叶子结点是0，第二层叶子结点有1个叶子结点）
参悟了前面几题有关树的题后，其实并不难
#include <iostream>
#include <vector>
using namespace std;
int n, m;
const int maxn = 110;
vector<int> child[maxn];
int ans[maxn];
int num = 0;
void DFS(int root, int depth)
{
if (child[root].size() == 0)
{
num = max(num, depth);
ans[depth]++;
return;
}
for (int i = 0; i < child[root].size(); i++)
{
DFS(child[root][i], depth + 1);
}
}
int main()
{
int id, temp,k;
cin >> n >> m;
for (int i = 1; i <= m; i++)
{
cin >> id >> k;
for (int j = 0; j < k; j++)
{
cin >> temp;
child[id].push_back(temp);
}
}
DFS(1, 1);
// cout << num << endl;
for (int i = 1; i <= num; i++)
{
cout << ans[i];
if (i != num)
cout << " ";
}
return 0;
}
