1094-- The Largest Generation (25)

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本文探讨了如何通过树的层次遍历算法，找到家族树中人口最多的代及其人数。通过对输入数据的解析和处理，利用队列进行层次遍历，并记录每一代的人口数量，最终找出人口最多的代。

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1005

题目：

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

算法分析：

树的层次遍历，求出拥有节点最多的层数及其节点数目。

（也可以记录每个节点的父亲，然后通过DFS递归求深度）

代码：

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

const int MAXN = 10001;
int N, M;
struct Tree {
	int level = 0;
	vector<int> kids;
}tree[MAXN];

void Input(void) {
	cin >> N >> M;
	for (int i = 1; i <= M; i++) {
		int index, n;
		cin >> index >> n;
		if (1 == index)
			tree[index].level = 1;
		for (int j = 1; j <= n; j++) {
			int temp;
			cin >> temp;
			tree[index].kids.push_back(temp);
		}
	}
}

//树的层次遍历
void LevelOrder(void) {
	queue<Tree> q;
	q.push(tree[1]);

	while (!q.empty()) {
		Tree temp = q.front();
		q.pop();
		for (int i = 0; i < temp.kids.size(); i++) {
			tree[temp.kids.at(i)].level = temp.level + 1;
			q.push(tree[temp.kids.at(i)]);
		}
	}
	return;
}

int main(void) {
	Input();
	LevelOrder();
	
	int level[101]; fill(level, level + 101, 0);
	int maxlevel = 0, maxcount = 0, levelindex = 0;
	for (int i = 1; i <= N; i++) {
		level[tree[i].level]++;
		if (maxlevel < tree[i].level)
			maxlevel = tree[i].level;
	}
	for (int i = 1; i <= maxlevel; i++)
		if (maxcount < level[i]) {
			maxcount = level[i];
			levelindex = i;
		}

	cout << maxcount << " " << levelindex << endl;

	system("pause");
	return 0;
}