题目

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

题目分析

已知每个节点的所有子节点，求节点最多的层数max_num_h及对应层结点数max_num

解题思路

思路 01（最优）

邻接表存储树，dfs深度优先遍历树（当前节点所在层级用函数参数记录），int cndn[n]记录每层节点数，并更新max_num和max_num_h

思路 02

邻接表存储树，bfs深度优先遍历树（当前节点所在层级用int h[n]记录），int cndn[n]记录每层节点数，并更新max_num和max_num_h

易错点

题目理解错误，求的是节点最多的层（而不是节点最多的分支），求的是节点最多的层（而不是叶结点最多的层）

Code

Code 01（dfs 最优）

#include <iostream>

#include <vector>

#include <queue>

using namespace std;

const int maxn=110;

vector<int> nds[maxn];

int cnds[maxn];//cnds[i] 记录i层结点总数

int max_num,max_num_h;//max_num树的宽度；max_num_h 结点数最多的层数；

void dfs(int now, int h) {

	cnds[h]++;

	if(max_num<cnds[h]) { //更新最多结点的层数信息

		max_num_h=h; //层最多结点数 （树的宽度）

		max_num=cnds[h]; //最多结点数的层号

	}

	if(nds[now].size()==0)return;

	for(int i=0; i<nds[now].size(); i++) { //遍历当前结点所有子结点，进行dfs遍历

		dfs(nds[now][i],h+1); // 当前结点的子结点所在层数=当前结点层数+1

	}

}

int main(int argc,char * argv[]) {

	int n,m,id,k,cid;

	scanf("%d %d",&n,&m); //n结点总数；m非叶子结点数

	for(int i=0; i<m; i++) { //输入所有非叶子结点的所有子节点信息

		scanf("%d %d",&id,&k); //id 父结点编号；k该父结点有多少个子结点

		for(int j=0; j<k; j++) {

			scanf("%d", &cid);

			nds[id].push_back(cid); //保存每个孩子到其父结点的容器中

		}

	}

	dfs(1,1);

	printf("%d %d",max_num,max_num_h);

	return 0;

}

Code 02（bfs）

#include <iostream>

#include <vector>

#include <queue>

using namespace std;

const int maxn=110;

vector<int> nds[maxn];

int h[maxn];//h[i] 记录i结点所在层数

int cnds[maxn];//cnds[i] 记录i层结点总数

int max_num,max_num_h;//max_num树的宽度；max_num_h 结点数最多的层数；

void bfs() {

	queue<int> q;//bfs需要借助队列实现

	q.push(1); //根结点编号为1

	while(!q.empty()) {

		int now = q.front();

		q.pop();

		cnds[h[now]]++;

		if(max_num<cnds[h[now]]) { //更新最多结点的层数信息

			max_num_h=h[now]; //层最多结点数 （树的宽度）

			max_num=cnds[h[now]]; //最多结点数的层号

		}

		for(int i=0; i<nds[now].size(); i++) { //遍历当前结点所有子结点，进行bfs遍历

			h[nds[now][i]]=h[now]+1; // 当前结点的子结点所在层数=当前结点层数+1

			q.push(nds[now][i]);

		}

	}

}

int main(int argc,char * argv[]) {

	int n,m,id,k,cid;

	scanf("%d %d",&n,&m); //n结点总数；m非叶子结点数

	for(int i=0; i<m; i++) { //输入所有非叶子结点的所有子节点信息

		scanf("%d %d",&id,&k); //id 父结点编号；k该父结点有多少个子结点

		for(int j=0; j<k; j++) {

			scanf("%d", &cid);

			nds[id].push_back(cid); //保存每个孩子到其父结点的容器中

		}

	}

	h[1]=1;

	bfs();

	printf("%d %d",max_num,max_num_h);

	return 0;

}

巴特西

PAT Advanced 1094 The Largest Generation (25) [BFS，DFS，树的遍历]

题目