PAT A1094

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

下面写了DFS和BFS两种做法,选择一项即可

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#include "stdio.h"
#include "math.h"
#include "string.h"
//#include "iostream"
//#include "stdlib.h"
#include "vector"
//#include "set"
//#include "map"
//#include "stack"
#include "queue"
#include "algorithm"
using namespace std;

//typedef long long LL;
const int maxn = 110;
int n, m;
struct node{
vector<int> child;
int level;
}Node[maxn];
int hashTable[maxn] = {0};
void BFS(int root){
queue<int> q;
Node[root].level = 1;
q.push(root);
while (!q.empty()) {
int now = q.front();
q.pop();
hashTable[Node[now].level]++;
for (int i = 0; i < Node[now].child.size(); i++) {
q.push(Node[now].child[i]);
Node[Node[now].child[i]].level = Node[now].level + 1;
}
}
}
void DFS(int index, int depth){
hashTable[depth]++;
for (int j = 0; j < Node[index].child.size(); j++) {
DFS(Node[index].child[j], depth + 1);
}
}

int main(){
scanf("%d %d", &n, &m);
for (int i = 0 ; i < m; i++) {
int id, k;
scanf("%d%d", &id, &k);
for (int j = 0; j < k; j++) {
int child;
scanf("%d", &child);
Node[id].child.push_back(child);
}
}
// BFS(1);
DFS(1, 1);
int maxk, maxnum = 0;
for (int i = 0; i < maxn; i++) {
if (hashTable[i] > maxnum) {
maxnum = hashTable[i];
maxk = i;
}
}
printf("%d %d", maxnum, maxk);
return 0;
}