1094 The Largest Generation

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 (<) 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 (<) 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 (>) 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

题意：

给出一个家谱图，找出这个家谱图中，人数最多的一代，以及这一代是第几代。

思路：

用map将父代与子代之间进行映射，然后进行层次遍历。层次遍历的时候，可以增加哨兵，方便记录层数。

Code：

#include<iostream>
#include<vector>
#include<map>
#include<queue>

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;

    map<int, vector<int> > mp;
    vector<int> son;
    for (int i = 0; i < m; ++i) {
        int f, cnt, s;
        cin >> f >> cnt;
        for (int j = 0; j < cnt; ++j) {
            cin >> s;
            son.push_back(s);
        }
        mp[f] = son;
        son.clear();
    }

    int count = 1, level = 1;
    int temp1 = 0, temp2 = 1;
    queue<int> q;
    q.push(1);
    q.push(0);
    while (q.size() > 1) {
        int f = q.front();
        q.pop();
        if (f == 0) {
            q.push(0);
            temp2 += 1;
            level = temp1 > count ? temp2 : level;
            count = temp1 > count ? temp1 : count;
            temp1 = 0;
        }
        if (mp.find(f) != mp.end()) {
            son = mp.find(f)->second;
            for (int i = 0; i < son.size(); ++i) {
                q.push(son[i]);
            }
            temp1 += son.size();
        }
            
    }

    cout << count << " " << level << endl;

    return 0;
}

　　

参考：

https://blog.csdn.net/xiaolonggezte/article/details/80004392