PAT甲级——1094 The Largest Generation (树的遍历)

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1094 The Largest Generation (25 分)

A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population.

Input Specification:

Each input file contains one test case. Each case starts with two positive integers

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

where ID is a two-digit number representing a family member, K ( $>) is the number of his/her children, followed by a sequence of two-digit ID's of his/her children. For the sake of simplicity, let us fix the root ID to be 01 . All the numbers in a line are separated by a space.$

Output Specification:

For each test case, print in one line the largest population number and the level of the corresponding generation. It is assumed that such a generation is unique, and the root level is defined to be 1.

Sample Input:

23 13
21 1 23
01 4 03 02 04 05
03 3 06 07 08
06 2 12 13
13 1 21
08 2 15 16
02 2 09 10
11 2 19 20
17 1 22
05 1 11
07 1 14
09 1 17
10 1 18

Sample Output:

9 4

题目大意：一个家族里面的人的ID编号是从01到N，每个人有若干个孩子，形成一棵族谱树，同一层的人是同辈关系，找出人数最多的那一层并且输出该层的人数。

思路：层序遍历（BFS思想）标记每个人所在的层数，另开一个数组 ans[level] = num 用于映射层数和人数的关系，树遍历完成后再遍历ans数组就能找到人数最多的那一层。

 1 #include <iostream>
 2 #include <vector>
 3 #include <queue>
 4 using namespace std;
 5  
 6 struct node {
 7     int level = 0;
 8     vector<int> child;
 9 };
10 vector<node> tree;
11 vector<int> ans;
12 void getLevel();
13 int main()
14 {
15     int N, M, ID, K;
16     scanf("%d%d", &N, &M);
17     tree.resize(N + 1);
18     ans.resize(N + 1, 0);
19     for (int i = 0; i < M; i++) {
20         scanf("%d%d", &ID, &K);
21         tree[ID].child.resize(K);
22         for (int j = 0; j < K; j++)
23             scanf("%d", &tree[ID].child[j]);
24     }
25     getLevel();
26     int level = 0, num = 0;
27     for (int i = 0; i < ans.size(); i++) {
28         if (num < ans[i]) {
29             num = ans[i];
30             level = i;
31         }
32     }
33     printf("%d %d\n", num, level);
34     return 0;
35 }
36  
37 void getLevel() {
38     int ID = 1;
39     queue<int> Q;
40     tree[ID].level = 1;
41     Q.push(ID);
42     while (!Q.empty()) {
43         ID = Q.front();
44         ans[tree[ID].level]++;
45         Q.pop();
46         int childID;
47         for (int i = 0; i < tree[ID].child.size(); i++) {
48             childID = tree[ID].child[i];
49             tree[childID].level = tree[ID].level + 1;
50             Q.push(childID);
51         }
52     }
53 }