pat1094 The Largest Generation-优快云博客

本文介绍了一种通过构建家族成员的树形结构来确定人口最多的代群的方法。该方法使用广度优先搜索算法遍历树形结构，并统计每一代的人口数量，最终找到人口最多的代及其对应的层级。

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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

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;

void BFS(vector<vector<int> > vec, int root) {
	queue<int> q;
	q.push(root);
	q.push(0);

	int cur = 0, maxx = 0, generation=0, gene=0;

	while(!q.empty()) {
		int tmp = q.front();
		
		q.pop();

		if(tmp == 0 && q.size() > 1) {
			generation += 1;
			//cout << "generation = " << generation << " " << "cur = " << cur <<endl;
			if (maxx < cur) {
				maxx = cur;
				gene = generation;
			}
			cur = 0;
			q.push(0);
			continue;
		} else if(tmp == 0 && q.size() <= 1) {
			generation += 1;
			//cout << "generation = " << generation << " " << "cur = " << cur <<endl;
			if (maxx < cur) {
				maxx = cur;
				gene = generation;
			}
			break;
		}
		cur += 1;
		vector<int> sub = vec[tmp];
		for(int i = 0; i < sub.size(); i++) {
			q.push(sub[i]);
		}
	}

	printf("%d %d\n", maxx, gene);
}

int main() {

	int n, m;
	int flag[110];

	scanf("%d %d", &n, &m);

	for (int i = 0; i <= n; ++i)
	{
		flag[i] = 0;
	}

	vector<vector<int> > trees(n+1);

	for (int i = 0; i < m; ++i)
	{
		int node, cnt, a;

		scanf("%d %d", &node, &cnt);

		for (int j = 0; j < cnt; ++j)
		{
			scanf("%d", &a);
			trees[node].push_back(a);
			flag[a] = 1;
		}
	}

	int root = 0;

	for (int i = 1; i <= n; i++) {
		if(flag[i] == 0) {
			root = i;
			break;
		}
	}
	
	BFS(trees, root);

	return 0;
}

如果使用cout，cin进行输出输入，会超时。