1004 Counting Leaves (30 分)

标签：orm   size   tps   seq   ring   cout   一个   show   cstring   

题目链接：

https://pintia.cn/problem-sets/994805342720868352/problems/994805521431773184

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.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0, the number of nodes in a tree, and M (<), 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

题目大意：

一个树，求其每层的叶子结点个数。

思路：

1、通过结构体存储树，将父节点记录。

2、知道了自己的父亲结点的同时也知道了自己的层数（父节点层数+1）。

3、自己是否有孩子结点在输入的时候就可以进行记录（ID即为非叶节点）。

注意事项：

1、输入并非有序，先把所有结点的父节点记录下来，不要直接记录层数，因为可能父节点还没有读取到，就会导致没有层数。

2、当只有一个根时，输出1，根也是一个结点（好吧，是我的树知识不过关~-~）

AC代码：

#include <iostream>
#include <cstring>

using namespace std;

struct
{
    int childHave = 0;    //是否为叶子节点
    int level;            //层数
    int father;           //父节点
}root[105];

int sum[105];

int main()
{
    memset(sum,0,sizeof(sum));
    int n,m;
    cin >> n >> m;          //记录数据
    root[1].level = 1;
    for(int i=1;i<=m;i++)
    {
        int r,k;
        cin >> r >> k;
        root[r].childHave = 1;
        for(int j=1;j<=k;j++)
        {
            int temp;
            cin >> temp;
            root[temp].father = r;
        }
    }
    int maxa = 0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(root[j].father == i) root[j].level = root[i].level+1;        //如果父亲节点为i则层数为i层数+1
            if(root[j].level > maxa) maxa = root[j].level;                  //更新最大层数
        }
    }
    for(int i=1;i<=n;i++) if(root[i].childHave == 0) sum[root[i].level]++;
    cout << sum[1];
    for(int i=2;i<=maxa;i++) cout << ‘ ‘ << sum[i];
    return 0;
}

View Code

1004 Counting Leaves (30 分)

标签：orm   size   tps   seq   ring   cout   一个   show   cstring   

原文地址：https://www.cnblogs.com/abszse/p/11919734.html