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

代码长度限制
16 KB
时间限制
200 ms
内存限制
64 MB

代码实现：

#include<iostream>
#include<queue>
#include<cstring>
#include<vector>
using namespace std;
const int N=105;
vector<int>v[N];
int level[N];
void bfs(int x){
    queue<pair<int,int> >q;
    q.push({1,x});
    while(q.size()){
        int dis=q.front().first;
        int s=q.front().second;
        q.pop();
        level[dis]++;
        for(int i=0;i<v[s].size();i++){
            q.push({dis+1,v[s][i]});
        }
    }
}
int main(){
    int n,m;
    cin>>n>>m;
    while(m--){
        string s;
        cin>>s;
        int k;
        cin>>k;
        while(k--){
            string s1;
            cin>>s1;
            v[stoi(s)].push_back(stoi(s1));
        }
    }
    bfs(1);
    int max1=0;
    for(int i=1;i<=n;i++)max1=max(max1,level[i]);
    for(int i=1;i<=n;i++){
        if(level[i]==max1){
            cout<<level[i]<<" "<<i<<endl;
            break;
        }
    }
    return 0;
}