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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 <cmath>
using namespace std;
vector<int> nds[101];//id从01开始
int cndn[101],max_num_h,max_num;
void dfs(int index, int h) {
    cndn[h]++;
    if(max_num<cndn[h]) {
        max_num=cndn[h];
        max_num_h=h;
    }
    if(nds[index].size()==0) return;
    for(int i=0; i<nds[index].size(); i++)
        dfs(nds[index][i],h+1);
}
int main(int argc,char * argv[]) {
    int n,m,id,cn,cid;
    scanf("%d %d",&n,&m);
    for(int i=0; i<m; i++) {
        scanf("%d %d", &id,&cn);
        for(int j=0; j<cn; j++) {
            scanf("%d", &cid);
            nds[id].push_back(cid);
        }
    }
    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=101;
vector<int> nds[maxn];
int cndn[maxn],max_num_h,max_num,h[maxn];
void bfs() {
    queue<int> q;
    q.push(1);
    while(!q.empty()) {
        int now = q.front();
        q.pop();
        cndn[h[now]]++;
        if(max_num<cndn[h[now]]) {
            max_num=cndn[h[now]];
            max_num_h=h[now];
        }
        for(int i=0; i<nds[now].size(); i++) {
            h[nds[now][i]]=h[now]+1;
            q.push(nds[now][i]);
        }
    }
}
int main(int argc,char * argv[]) {
    int n,m,id,cn,cid;
    scanf("%d %d",&n,&m);
    for(int i=0; i<m; i++) {
        scanf("%d %d",&id,&cn);
        for(int j=0; j<cn; j++) {
            scanf("%d",&cid);
            nds[id].push_back(cid);
        }
    }
    h[1]=1;
    bfs();
    printf("%d %d",max_num,max_num_h);
    return 0;
}

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

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原文地址：https://www.cnblogs.com/houzm/p/12325246.html