A1094

1094 The Largest Generation (25分)

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

把树画出来，就很直观地理解问题。

#include<bits/stdc++.h>
using namespace std;
struct node{
    vector<int> child;
}tree[1010];
int N,M,sum,l=0;
void bfs(int root,int l,int k){
    if(l==k){
        ++sum;
        return ;
    }
    for(int i=0;i<tree[root].child.size();i++)
        bfs(tree[root].child[i],l+1,k);
}
int main(){
    cin>>N>>M;
    for(int i=0;i<M;i++){
        int x,n,data;
        scanf("%d %d",&x,&n);
        for(int j=0;j<n;j++){
            scanf("%d",&data);
            tree[x].child.push_back(data);
        }
    }
    int max=0,index=0;
    for(int i=1;i<50;i++){
        sum=0,l=1;
        bfs(1,l,i);
        if(sum>max){
            max=sum;
            index=i;
        }
    }
    cout<<max<<' '<<index<<endl;
    return 0;
}