1094 The Largest Generation （25 point(s))

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此算法解析家族树结构，通过深度优先搜索确定每代人口数量，旨在找出人数最多的一代及其层级，适用于理解家族成员分布。

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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 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：第一行给出树的总结点个数n和非叶子结点个数m，下面m行每一行给出非叶子结点编号，孩子个数k，k个孩子编号id【】。输出最宽的一层结点个数max以及层号layer。（根节点编号为01，属于第一层）

分析：由于每一行给出的是非叶子结点与孩子结点（只有父子关系，并无左右孩子关系，不适用二叉链表存储），把树看作图，采用邻接表存储，从根节点开始dfs深搜统计每一层节点个数sum【depth】，最后循环比较sum【i】，输出最大sum【i】和 i。

核心代码：

int maxdepth=-1;
void dfs(int index,int depth){
	sum[depth]++;
	if(v[index].size()==0){
		maxdepth=max(depth,maxdepth)+1;
		return ;
	}
	
	for(int i=0;i<v[index].size();i++){
		dfs(v[index][i],depth+1);
	}
}

对比，若采用二叉链表存储，可改为

int maxdepth = -1;
void DFS(node *root, int depth) {
    if(!root) {
        maxdepth = max(depth, maxdepth);
        return ;
    }
	num[depth]++;
  //  if(!root) return ;
    DFS(root->left, depth + 1);
    DFS(root->right, depth + 1);
}

完整代码：

#include<bits/stdc++.h>
using namespace std;
//struct node{
//	int data;
//	int layer;
//	vector<int> child;
//}; node *Node[110];
vector<int> v[110];
int maxdepth=-1;
int sum[110];
void dfs(int index,int depth){
	sum[depth]++;
	if(v[index].size()==0){
		maxdepth=max(depth,maxdepth)+1;
		return ;
	}
	
	for(int i=0;i<v[index].size();i++){
		dfs(v[index][i],depth+1);
	}
}

int main(){
	int n,m,k,id,val;
	cin>>n>>m;
	for(int i=0;i<m;i++){
		cin>>id>>k;
		for(int i=0;i<k;i++){
			cin>>val;
			v[id].push_back(val);
		}
	}
	dfs(1,1);
	int max=-1,layer=1;
	for(int i=0;i<maxdepth;i++){
		if(sum[i]>max){
			max=sum[i];
			layer=i;
		}
	}
	cout<<max<<" "<<layer<<endl;
	return 0;
}

上一题：1087 All Roads Lead to Rome （30 point(s))

https://blog.youkuaiyun.com/qq_41317652/article/details/89139258