PAT树的深度优先搜索（dfs）和广度优先搜索（bfs）
dfs和bfs都是常用的搜索算法，结合PAT1094，简单动手分别以dfs和bfs实现树的遍历吧。
1094 The Largest Generation (25 分)
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]whereIDis a two-digit number representing a family member,K(>0) is the number of his/her children, followed by a sequence of two-digitID's of his/her children. For the sake of simplicity, let us fix the rootIDto be01. 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 18Sample Output:
9 4题目大意：给你一棵树，让你找出哪一层的节点个数最多，输出节点个数和对应的层数。
dfs：
#include<cstdio> #include<iostream> #include<vector> const int maxn=105; vector<int> tree[maxn]; int sum[maxn];//记录每一层的节点个数 int n,m; void dfs(int level,int node){ sum[level]++; for(int i=0;i<tree[node].size();i++){ dfs(level+1,tree[node][i]); } } int main() { cin>>n>>m; for(int i=0;i<m;i++){ int id,num; cin>>id>>num; for(int j=0;j<num;j++){ int child; cin>>child; tree[id].push_back(child); } } dfs(1,1); int maxnum=-1,maxlevel=-1; for(int i=0;i<maxn;i++){ if(sum[i]>maxnum){ maxnum=sum[i]; maxlevel=i; } } cout<<maxnum<<" "<<maxlevel<<endl; return 0; }bfs
#include<cstdio> #include<iostream> #include<vector> #include<queue> using namespace std; const int maxn=105; vector<int> tree[maxn]; int sum[maxn]; int level[maxn]; //记录节点对应的层数 int n,m; void bfs(){ queue<int> q; q.push(1); level[1]=1; while(!q.empty()){ int node=q.front(); q.pop(); sum[level[node]]++; for(int i=0;i<tree[node].size();i++){ level[tree[node][i]]=level[node]+1; q.push(tree[node][i]); } } } int main() { cin>>n>>m; for(int i=0;i<m;i++){ int id,num; cin>>id>>num; for(int j=0;j<num;j++){ int child; cin>>child; tree[id].push_back(child); } } bfs(); int maxnum=-1,maxlevel=-1; for(int i=0;i<maxn;i++){ if(sum[i]>maxnum){ maxnum=sum[i]; maxlevel=i; } } cout<<maxnum<<" "<<maxlevel<<endl; return 0; }