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