1094. The Largest Generation(25) 이 문항은 6도 공간과 같은 이치로 BFS의 차원을 기록한다.

3245 단어 bfs
1094. The Largest Generation (25)
시간 제한
200 ms
메모리 제한
65536 kB
코드 길이 제한
16000 B
판정 절차
Standard
작자
CHEN, Yue
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 (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
//동창이 있으면 찾아서 보내도 된다
//6도 공간http://blog.csdn.net/u013167299/article/details/42321615
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<queue>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=105;
const int inf=210000;
vector<int> g[maxn];
bool vis[maxn];
int lev=1;
int sum;
int ans=1;//ans=0          ,       ,     
void bfs(int s)
{
    queue<int> q;
    q.push(s);
    vis[s]=true;
    int tail;
    int last=s;
    int cnt=1;
    while(!q.empty())
    {
        int k=q.front();
        q.pop();
        int n=g[k].size();
        for(int i=0;i<n;i++)
        {
            if(!vis[g[k][i]])
            {
                q.push(g[k][i]);
                vis[g[k][i]]=true;
                tail=g[k][i];
                sum++;
            }
        }
        if(k==last)
        {
            last=tail;
            cnt++;
            if(ans<sum)
            {
                ans=sum;
                lev=cnt;
            }
            sum=0;
        }
    }
}

int main()
{
    int n,m,i,j,k,t;
    freopen("in.txt","r",stdin);
    scanf("%d%d",&n,&m);
    for(i=0;i<m;i++)
    {
        int id;
        cin>>id>>k;
        for(j=0;j<k;j++)
        {
            cin>>t;
            g[id].push_back(t);
            g[t].push_back(id);
        }
    }
    bfs(1);
    printf("%d %d
",ans,lev); return 0; }

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