1004 Counting Leaves

最新推荐文章于 2022-07-26 16:28:25 发布

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本文介绍了一种算法，用于解决家庭树中无子女成员的计数问题。输入包括节点数量和非叶节点详情，输出为各层级的叶节点数量。通过递归遍历树结构，实现了对叶节点的有效统计。

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 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

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
struct Node
{
    int val;
    vector<int> child;
}node[101];
int n,m,maxl=-1;
int ans[101]={0};
void findans(int root,int lay)
{
    maxl=max(maxl,lay);
    if(node[root].child.size()==0)///叶子
    {
        ans[lay]++;
        return;
    }
    for(int i=0;i<node[root].child.size();i++)
        findans(node[root].child[i],lay+1);
}
int main()
{
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        int id,k,e;
        cin>>id>>k;
        for(int j=0;j<k;j++)
        {
            cin>>e;
            node[id].child.push_back(e);
        }
    }
    findans(1,0);
    cout<<ans[0];
    for(int i=1;i<maxl+1;i++)
        cout<<" "<<ans[i];
    cout<<endl;
    return 0;
}