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结尾无空行

Sample Output:

9 4结尾无空行

思路

这道题和1004非常相似， 一般树的遍历。信息多时用结构体存，只有子节点是vector数组即可。

代码

#include <bits/stdc++.h>
using namespace std;

const int maxn = 105;
vector<int> tree[maxn];
int cnt[maxn];
int maxid = 0, maxnum = 0;
int n, m;

void dfs (int root, int level) {
    cnt[level]++;
    if (maxnum < cnt[level]) {
        maxnum = cnt[level];
        maxid = level;
    }
    for (int i = 0; i < tree[root].size(); i++) {
        dfs(tree[root][i], level + 1);
    }
}

int main() {
    cin>>n>>m;
    for (int i = 0; i < m; i++) {
        int a, len;
        cin>>a>>len;
        for (int j = 0; j < len; j++) {
            int num;
            cin>>num;
            tree[a].push_back(num);
        }
    }
    dfs(1, 1);
    cout<<maxnum<<" "<<maxid<<endl;
}