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;
}