The Largest Generation (25)

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.

输入例子:

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

输出例子:

9 4

#include<bits/stdc++.h>
using namespace std;
map<int,vector<int>> tree;
queue<int> root;
int level=1;
int max_level=1;
int max_population=-1;
queue<int> next_level(queue<int> q){ //层序遍历
    queue<int> child;
    level++;
    if(!q.empty()){
        int temp;
        while(!q.empty()){
            temp=q.front();
            vector<int>::iterator it;
            for (it=tree[temp].begin();it!=tree[temp].end();++it){
                child.push(*it);
            }
            q.pop();
        }
        if (max_population<child.size()){
            max_population=child.size();
            max_level=level+1;
        }
        return next_level(child);
    }
    return child;
}
int main(){
    int n,m,k;
    int id;
    cin>>n>>m;
    for (int i=1;i<=m;++i){
        cin>>id>>k;
        int temp;
        for (int j=1;j<=k;++j){
            cin>>temp;
            tree[id].push_back(temp);
            if (id==1){
                root.push(temp);
            }
        }
    }
    max_population=root.size();
    next_level(root);
    cout<<max_population<<" "<<max_level<<endl;
}

题目翻译一下就是找一棵树中兄弟最多的一层，通过层序遍历即可直接实现