甲级1004 Counting Leaves

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


题意：一个家族的层次关系可由血缘树表示。输入家族总人数N和非叶子节点的人数M（非叶子节点即有子女的人），之后M行输入非叶子节点的子节点。最后求出每一层级中，叶子节点的数量
分析：一开始想到的是构建一个树并层序遍历，然后输出每一层的叶子节点数量。这个做法写完后有两个点就超时了。
看了一眼其他人的想法，其中提到一句：每个节点的子节点对其影响不大，重要的是其父节点。对于这道题确实是这样，父节点可以推导出当前层级。而子节点则无法提供任何有用信息。
所以每个节点只存储父节点，层级，是否有子节点的信息。
整个程序先读入所有信息，将所有节点的父节点确定下来，然后再求每个节点的层级，这个过程用递归实现，实际上由于我没有用vector来存储各个节点的信息，而是用一个已经开好的数组然后用ID做下标访问信息，所以这里求各个节点层级的速度我认为应该相当快。
最后一步就是遍历所有节点的信息，得到结果。
坑点：1 0的测试样例输出必须为1。

AC代码如下：

#include <iostream>
using namespace std;

struct Node
{
    int parent;
    int level;        //parent的层级+1
    bool haveChildren;
    bool isFamily;
};

Node family[100];
int levelLeaf[105];
int maxLevel = 0;

int getlevel(int ID)
{
    if (ID == 1)    return 0;
    if (family[ID].level == 0)
    {
        family[ID].level = getlevel(family[ID].parent) + 1;
        if (family[ID].level > maxLevel)    maxLevel = family[ID].level;
    }
    return family[ID].level;
}

int main()
{    
    int N, M;
    scanf("%d", &N);
    if (N == 0)    return 0;
    scanf("%d", &M);
    int ID_in;
    family[1].isFamily = true;
    for (int i = 0; i < M; i++)
    {
        scanf("%d", &ID_in);
        family[ID_in].haveChildren = true;
        family[ID_in].isFamily = true;
        int child_count,child_ID;
        scanf("%d", &child_count);
        for (int j = 0; j < child_count; j++)
        {
            scanf("%d", &child_ID);
            family[child_ID].parent = ID_in;
            family[child_ID].isFamily = true;
        }
    }
    for (int i = 0; i < 100; i++)
    {
        if (family[i].isFamily == true && family[i].haveChildren == false)
        {
            levelLeaf[getlevel(i)]++;
        }
    }
    bool flag = false;
    for (int i = 0; i <= maxLevel; i++)
    {
        if (flag == false)    flag = true;
        else printf(" ");
        printf("%d", levelLeaf[i]);
    }
    return 0;
}

然后是两个点有超时，其它点通过的代码（如果有大神发现问题，请不吝赐教）：

#include <iostream>
#include <vector>
#include <unordered_map>
#include <string>
#include <queue>
using namespace std;

struct Node
{
    string ID;
    vector<string> children;
    int level;
};

unordered_map<string, Node*> mapp;
queue<string> analyse_que;

int main()
{
    /*N为所有结点数，M为非叶子节点数，求每一层的叶子节点数*/
    int N, M;
    cin >> N >> M;
    string ID_in;
    int children_count;
    Node root;
    root.ID = "01";
    root.level = 0;
    mapp["01"] = &root;
    for (int i = 0; i < M; i++)
    {
        cin >> ID_in;
        auto it = mapp.find(ID_in);
        if (it == mapp.end())
        {
            Node node;
            mapp[ID_in] = &node;
            node.ID = ID_in;
        }
        it = mapp.find(ID_in);
        cin >> children_count;
        string child_ID;
        for (int j = 0; j < children_count; j++)
        {
            cin >> child_ID;
            (it->second)->children.push_back(child_ID);
            auto child_it = mapp.find(child_ID);
            if (child_it == mapp.end())
            {
                //Node node;此处不能声明一个局部变量
                Node* node_p = new Node;
                mapp[child_ID] = node_p;
                node_p->ID = child_ID;
            }
        }
    }
    int non_leaf_count = 0;
    analyse_que.push("01");
    int leaf = 0;
    int now_level = 0;
    while (!analyse_que.empty())
    {
        string id = analyse_que.front();
        analyse_que.pop();
        Node* p = mapp[id];
        if ((p->children).empty())
        {
            if (p->level != now_level)
            {
                printf("%d ", leaf);
                for (int i = 0; i < (p->level - now_level - 1); i++)
                    printf("0 ");
                leaf = 1;
                now_level=p->level;
            }
            else
            {
                leaf++;
            }
        }
        else
        {
            for (auto it = (p->children).begin(); it != (p->children).end(); it++)
            {
                Node* child_p = mapp[*it];
                child_p->level = p->level + 1;
                analyse_que.push(child_p->ID);
            }
        }
    }
    printf("%d", leaf);
    return 0;
}