## Pat(A) 1094. The Largest Generation (25)

### 1094. The Largest Generation (25)

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

#include "iostream"
#include "vector"
#include "queue"
using namespace std;

int N,M;
vector<vector<int>> G;
int level,ans;

void init(){
cin>>N>>M;
G.resize(N + 1);
int i,k,j,x;
for(i = 0; i < M; i++){
cin>>x>>k;
G[x].resize(k);
for(j = 0; j < k; j++){
cin>>G[x][j];
}
}
}

void bfs(){
int root = 1;
int x,i;
queue<int> que;
level = 1;
ans = 1;

int numInLevel = 0,numInNextLevel = 0,num = 0,tempLevel = 0;

que.push(root);
numInLevel ++;
numInNextLevel;
tempLevel ++;
while(!que.empty()){
x = que.front();
que.pop();
num ++;
for(i = 0; i < G[x].size(); i++){
que.push(G[x][i]);
numInNextLevel ++;
}
if(num == numInLevel){
if(numInNextLevel > ans){
level = tempLevel + 1;
ans = numInNextLevel;
}
tempLevel ++;
numInLevel = numInNextLevel;
numInNextLevel = 0;
num = 0;
}
}

}

int main(){
init();
bfs();
printf("%d %d",ans,level);
//system("pause");
}

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