c++PTA A1094(vector+dfs)

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.

核心思路

这道题用vector存储孩子节点数量，只要孩子结点一直有那就不断地DFS，最后遍历输出就行。

完整源码

#include<cstdio>
#include<vector>
using namespace std;
const int MAXN = 110;
vector<int> Node[MAXN]; //树的静态写法,Node[i]存放结点i的孩子结点编号
int ha[MAXN] = {0}; //记录每层的结点个数
void DFS(int index,int level){
    ha[level]++;
    for(int j = 0;j<Node[index].size();j++){
        DFS(Node[index][j],level+1); //遍历所有孩子结点，进行递归
    }
}
int main()
{
    int n,m,parent,k,child;
    scanf("%d%d",&n,&m);
    for(int i =0;i<m;i++){
        scanf("%d%d",&parent,&k); //父亲结点编号 ，孩子个数
        for(int j = 0;j<k;j++){
            scanf("%d",&child); //孩子结点编号
            Node[parent].push_back(child); // 建树
        }
    }
    DFS(1,1); //根节点为1，层号为1
    int maxLevel = -1,maxValue = 0; 
    for(int i =1;i<MAXN;i++){ //计算ha数组最大值
        if(ha[i]>maxValue){
            maxValue = ha[i];
            maxLevel = i;
        }
    }
    printf("%d %d\n",maxValue,maxLevel); // 输出最大的结点数与盖层层号
    return 0;
}