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

分析

設(shè)置數(shù)組dep存儲每個depth下節(jié)點(diǎn)個數(shù)华糖，dfs遍歷一遍樹麦向，再在dep中找到最大值和所在位置即可

code

#include <iostream>
#include <vector>

using namespace std;

struct node{
    vector<int> child;
    int depth;
};

vector<node> v;
vector<int> dep;
void dfs(int root,int depth){
    v[root].depth = depth;
    dep[depth]+=1;
    if(v[root].child.size()==0){
        return ;
    }
    for(int i=0;i<v[root].child.size();i++){
        dfs(v[root].child[i],depth+1);
    }
}



int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    v.resize(n+1);
    dep.resize(n+1);
    int a,b,c;
    for(int i=0;i<m;i++){
        scanf("%d %d",&a,&b);
        for(int j=0;j<b;j++){
            scanf("%d",&c);
            v[a].child.push_back(c);
        }
    }
    dfs(1,1);
    int maxd=1,maxg=1;
    for(int i=1;i<n;i++){
        if(dep[i]>maxd){
            maxd = dep[i];
            maxg = i;
        }
    }
    printf("%d %d",maxd,maxg);
    return 0;
}