The Largest Generation

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题目描述:

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.

输入描述:

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.

输出描述:

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.

输入样例:

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

输出样例:

9 4

本人AC代码:

#include<bits/stdc++.h>
using namespace std;
struct TreeNode{int val;vector<TreeNode*> son;TreeNode(int x):val(x),son({}){};
};
int main(){ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int n,m;cin>>n>>m;vector<TreeNode*> tree(n);for(int i=0;i<n;i++){TreeNode* node=new TreeNode(0);tree[i]=node;}for(int i=0;i<m;i++){int f;int k;cin>>f>>k;for(int j=0;j<k;j++){int p;cin>>p;tree[f-1]->son.push_back(tree[p-1]);}}queue<TreeNode*> q;int num=1;int lev=1;int len=1;int ans=1;q.push(tree[0]);while(!q.empty()){TreeNode* node=q.front();q.pop();int n=node->son.size();for(int i=0;i<n;i++){q.push(node->son[i]);}if(--len==0){len=q.size();lev++;if(len>num){num=len;ans=lev;}}}cout<<num<<' '<<ans<<endl;return 0;
}