PAT Advanced 1094 The Largest Generation

1094 The Largest Generation

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    

解题思路

• dfs

  dfs 遍历整个二叉树， 到每层处，对应的结点对应的层数值加1 （ num[level]++）， 最后枚举一遍找到最大值以及对应的层数 。

• bfs

  bfs遍历整个二叉树， 每层节点结点的 层数等于 父节点 层数+ 1， num[v[head][i]] = num[head] + 1， 找到层数后， 对应节点的层数加一h[num[head]] ++

解题代码

dfs：递归到每层， 对应层数的结点个数加1

#include 
#include 
using namespace std;
const int  N = 110;
int maxn, maxl, num[N];
vector<int> v[N];
void dfs(int cur, int level){
     
    num[level]++;
    for (int i = 0; i < v[cur].size(); i++){
     
        dfs(v[cur][i], level + 1);
    }
}
int main(){
     
    int n, m, p, k, c;
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++){
     
        scanf("%d %d", &p, &k);
        for (int j =0; j < k; j++){
     
            scanf("%d", &c);
            v[p].push_back(c);
        }
    }
    dfs(1, 1);
    for (int i = 0; i < N; i++){
     
        if (num[i] > maxn) maxn = num[i], maxl = i;
    }
    printf("%d %d", maxn, maxl);
    return 0;
}

bfs: 首先num表示每天节点的层数，通过hash 计数没层结点的个数，两次hash的使用。

#include 
#include 
#include 
using namespace std;
const int  N = 100;
int maxn, maxl, num[N], h[N];
vector<int> v[N];
void bfs(){
     
    queue<int> q;
    num[1] = 1;
    q.push(1);
    while (!q.empty()){
     
        int head = q.front();
        q.pop();
        int n = v[head].size();
        for (int i = 0; i < n; i++)
            q.push(v[head][i]), num[v[head][i]] = num[head] + 1;
        h[num[head]] ++;
    }
}
int main(){
     
    int n, m, p, k, c;
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++){
     
        scanf("%d %d", &p, &k);
        for (int j =0; j < k; j++){
     
            scanf("%d", &c);
            v[p].push_back(c);
        }
    }
    bfs();
    for (int i = 0; i < N; i++)
        if (maxn < h[i]) maxn = h[i], maxl = i;
    printf("%d %d", maxn, maxl);
    return 0;
}