1094. The Largest Generation (25) 此题和六度空间一个道理，记录BFS的层次

1094. The Largest Generation (25)
时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
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 (<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

//有同学用并查集也可以过的

//六度空间http://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的话第二个样例过不去，能找到这个错误，我也是醉了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\n",ans,lev);    return 0;}