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

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;
}