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一、题目

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 N (<100) which is the total number of family members in the tree (and hence assume that all the members are numbered from 01 to N), and M (<N) which is the number of family members who have children. Then M lines follow, each contains the information of a family member in the following format:

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

where ID is a two-digit number representing a family member, K (>0) 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 1321 1 2301 4 03 02 04 0503 3 06 07 0806 2 12 1313 1 2108 2 15 1602 2 09 1011 2 19 2017 1 2205 1 1107 1 1409 1 1710 1 18

Sample Output:

9 4

二、题目大意

输入树的结点个数N，结点编号为1~N，非叶子结点个数M，然后输出M个非叶子结点格子的孩子结点的编号，求结点个数最多的一层，根结点的层号为1，输出该层的结点个数以及层号。

三、考点

树、DFS

四、注意

1、使用struct node{}建树；

2、使用vector数组，在dfs的过程中维护每个depth的节点个数；

3、遍历vector找到最大的层。

五、代码

#include
   
    #include
    
     using namespace std;struct node {	vector
     
       v;};vector
      
        vec;vector
       
         vec_dep;void dfs(int root, int depth) {	vec_dep[depth]++;	//leaf	if (vec[root].v.size() == 0) {		return;	}	//next	for (int i = 0; i < vec[root].v.size(); ++i) {		dfs(vec[root].v[i], depth + 1);	}	return;}int main() {	//read	int n, m;	cin >> n >> m;	vec.resize(n + 1);	vec_dep.resize(n + 1, 0);	//read and build tree	while (m--) {		int id, k;		cin >> id >> k;		vec[id].v.resize(k);		for (int i = 0; i < k; ++i) {			cin >> vec[id].v[i];		}	}	//dfs	dfs(1, 1);	//max	int max_pop = 0, max_gen = 1;	for (int i = 1; i <= n; ++i) {		if (vec_dep[i] > max_pop) {			max_pop = vec_dep[i];			max_gen = i;		}	}	//output	cout << max_pop << " " << max_gen;	system("pause");	return 0;}
       
      
     
    
   

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