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：
 1 #include <iostream>
 2 using namespace std;
 3 
 4 struct Node
 5 {
 6     int parent;
 7     int level;        //parent的层级+1
 8     bool haveChildren;
 9     bool isFamily;
10 };
11 
12 Node family[100];
13 int levelLeaf[105];
14 int maxLevel = 0;
15 
16 int getlevel(int ID)
17 {
18     if (ID == 1)    return 0;
19     if (family[ID].level == 0)
20     {
21         family[ID].level = getlevel(family[ID].parent) + 1;
22         if (family[ID].level > maxLevel)    maxLevel = family[ID].level;
23     }
24     return family[ID].level;
25 }
26 
27 int main()
28 {    
29     int N, M;
30     scanf("%d", &N);
31     if (N == 0)    return 0;
32     scanf("%d", &M);
33     int ID_in;
34     family[1].isFamily = true;
35     for (int i = 0; i < M; i++)
36     {
37         scanf("%d", &ID_in);
38         family[ID_in].haveChildren = true;
39         family[ID_in].isFamily = true;
40         int child_count,child_ID;
41         scanf("%d", &child_count);
42         for (int j = 0; j < child_count; j++)
43         {
44             scanf("%d", &child_ID);
45             family[child_ID].parent = ID_in;
46             family[child_ID].isFamily = true;
47         }
48     }
49     for (int i = 0; i < 100; i++)
50     {
51         if (family[i].isFamily == true && family[i].haveChildren == false)
52         {
53             levelLeaf[getlevel(i)]++;
54         }
55     }
56     bool flag = false;
57     for (int i = 0; i <= maxLevel; i++)
58     {
59         if (flag == false)    flag = true;
60         else printf(" ");
61         printf("%d", levelLeaf[i]);
62     }
63     return 0;
64 }


 


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）：




 1 #include <iostream>
 2 #include <vector>
 3 #include <unordered_map>
 4 #include <string>
 5 #include <queue>
 6 using namespace std;
 7 
 8 struct Node
 9 {
10     string ID;
11     vector<string> children;
12     int level;
13 };
14 
15 unordered_map<string, Node*> mapp;
16 queue<string> analyse_que;
17 
18 int main()
19 {
20     /*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*/
21     int N, M;
22     cin >> N >> M;
23     string ID_in;
24     int children_count;
25     Node root;
26     root.ID = "01";
27     root.level = 0;
28     mapp["01"] = &root;
29     for (int i = 0; i < M; i++)
30     {
31         cin >> ID_in;
32         auto it = mapp.find(ID_in);
33         if (it == mapp.end())
34         {
35             Node node;
36             mapp[ID_in] = &node;
37             node.ID = ID_in;
38         }
39         it = mapp.find(ID_in);
40         cin >> children_count;
41         string child_ID;
42         for (int j = 0; j < children_count; j++)
43         {
44             cin >> child_ID;
45             (it->second)->children.push_back(child_ID);
46             auto child_it = mapp.find(child_ID);
47             if (child_it == mapp.end())
48             {
49                 //Node node;此处不能声明一个局部变量
50                 Node* node_p = new Node;
51                 mapp[child_ID] = node_p;
52                 node_p->ID = child_ID;
53             }
54         }
55     }
56     int non_leaf_count = 0;
57     analyse_que.push("01");
58     int leaf = 0;
59     int now_level = 0;
60     while (!analyse_que.empty())
61     {
62         string id = analyse_que.front();
63         analyse_que.pop();
64         Node* p = mapp[id];
65         if ((p->children).empty())
66         {
67             if (p->level != now_level)
68             {
69                 printf("%d ", leaf);
70                 for (int i = 0; i < (p->level - now_level - 1); i++)
71                     printf("0 ");
72                 leaf = 1;
73                 now_level=p->level;
74             }
75             else
76             {
77                 leaf++;
78             }
79         }
80         else
81         {
82             for (auto it = (p->children).begin(); it != (p->children).end(); it++)
83             {
84                 Node* child_p = mapp[*it];
85                 child_p->level = p->level + 1;
86                 analyse_que.push(child_p->ID);
87             }
88         }
89     }
90     printf("%d", leaf);
91     return 0;
92 }




 














            

            
posted @ 
2019-01-24 14:39 
囧YY 
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