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 树的层序遍历，统计结点最多的层以及该层上的结点数。使用队列BFS即可
#include <bits/stdc++.h>
#define MAXN 100+50
using namespace std;
int n, m;
vector<int> arr[MAXN];
int res[MAXN];
struct node{
int id;
int level;
};
int main(){
memset(res, 0, sizeof(res));
scanf("%d%d", &n, &m);
for(int i = 0; i < m; ++i){
int par, t, child;
scanf("%d%d", &par, &t);
for(int j = 0; j < t; ++j){
scanf("%d", &child);
arr[par].push_back(child);
}
}
queue<node> q;
q.push({1,1});
while(!q.empty()){
node top = q.front();
res[top.level]++;
q.pop();
for(int j = 0; j < arr[top.id].size(); ++j){
q.push({arr[top.id][j], top.level+1});
}
}
int maxNum = 0, k = 0;
for(int i = 1; i < MAXN; ++i){
if(res[i] > maxNum){
maxNum = res[i];
k = i;
}
}
cout << maxNum << " " << k << endl;
return 0;
}
CAPOUIS‘CODE