Counting Leaves

1004计数叶子节点问题解析

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博客围绕1004计数叶子节点问题展开，介绍了问题背景，即根据家谱树统计无子女家庭成员数量。给出输入格式，包含节点数和非叶子节点信息，还说明了输出要求，需按层级统计并输出无子女成员数量，最后给出示例输入输出。

1004 Counting Leaves (30)（30 分）

A family hierarchy is usually presented by a pedigree tree. Your job isto count those family members who have no child.

Input

Each input file contains one test case. Each case starts with a linecontaining 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 theformat:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K isthe number of its children, followed by a sequence of two-digit ID's ofits children. For the sake of simplicity, let us fix the root ID to be01.

Output

For each test case, you are supposed to count those family members whohave no child for every seniority level starting from the root. Thenumbers must be printed in a line, separated by a space, and there mustbe no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is theroot and 02 is its only child. Hence on the root 01 level, there is 0leaf node; and on the next level, there is 1 leaf node. Then we shouldoutput "0 1" in a line.

Sample Input

2 1
01 1 02

Sample Output

0 1

题目大意：从根节点开始算等级，输出每一等即没有孩子的家庭成员;

思路：先建树，用vis判断有无孩子，查找每个人所在的等级，并记为le[i],计算等级为i没有孩子的数目s[i];

代码如下:

#include<bits/stdc++.h>
using namespace std;
int pa[110],vis[110],le[110],s[110];//vis[0]代表没有孩子，le计算i的等级,s计算等级为i的没有孩子的数目;
int root(int r)
{
    int l=0;
    while(pa[r]!=r)
    {
        r=pa[r];
        l++;
    }
    return l;
}
int main()
{
    int n,m,maxn=0;
    cin>>n>>m;
    for(int i=0;i<=99;i++)
        pa[i]=i;
    memset(vis,0,sizeof(vis));
    while(m--)
    {
        int a,b;
        cin>>a;
        int k;
        cin>>k;
        while(k--)
        {
            cin>>b;
            pa[b]=a;
            vis[a]=1;
        }
    }
    for(int i=1;i<=n;i++)
    {
        le[i]=root(i);
        maxn=max(maxn,le[i]);
    }
    for(int i=1;i<=n;i++)
        if(!vis[i])
            s[le[i]]++;
    cout<<s[0];
    for(int i=1;i<=maxn;i++)
        cout<<" "<<s[i];
    cout<<endl;
    return 0;
}