树的遍历——A1094 The Largest Generation(25)（树的静态结构——> vector<int> Node[maxn]来表示）

法1：BFS

#include <bits/stdc++.h>
#include <stdio.h>
#include <stdlib.h>
#include <queue>
using namespace std;
const int maxn = 100010;
vector<int> temp[maxn];
int n,m;
int num,maxnum,level = 1,maxLevel = 1;
queue<int> que;
void BFS(){
    for(int i=0;i<temp[1].size();++i){
        que.push(temp[1][i]);
    }
    level++;
    num = que.size();
    maxnum = num;
    while(!que.empty()){
        for(int i = 0;i<num;++i){
            int digit = que.front();
            for(int j=0;j<temp[digit].size();++j){
                que.push(temp[digit][j]);
            }
            que.pop();
        }
        num = que.size();
        level++;
        if(que.size() > maxnum){
            maxnum = que.size();
            maxLevel = level;
        }
    }
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;++i){
        int father,number,son;
        scanf("%d%d",&father,&number);
        for(int j=0;j<number;++j){
            scanf("%d",&son);
            temp[father].push_back(son);
        }
    }
    num = temp[1].size();
    BFS();
    printf("%d %d
",maxnum,maxLevel);
    system("pause");
    return 0;
}

---

法2：DFS遍历

#include <bits/stdc++.h>
#include <stdio.h>
#include <stdlib.h>
#include <queue>
using namespace std;
const int maxn = 110;
vector<int>Node[maxn];
int hashTable[maxn] = {0};//记录每个层的节点数
void DFS(int index,int level){
    hashTable[level]++;
    for(int j = 0;j<Node[index].size();++j){
        DFS(Node[index][j],level+1);
    }
}
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);
        for(int j =0;j<k;++j){
            scanf("%d",&child);
            Node[parent].push_back(child);
        }
    }
    DFS(1,1);
    int maxLevel = -1,maxValue = 0;
    for(int i=1;i<maxn;++i){
        if(hashTable[i]>maxValue){
            maxValue = hashTable[i];
            maxLevel = i;
        }
    }
    printf("%d %d
",maxValue,maxLevel);
    system("pause");
    return 0;
}