20231124

/*
time:O(unknown)
space:O(n*n)
knowledge:树的直径 
step:dfs*2 
*/
#include<bits/stdc++.h>
using namespace std;
int n;
vector<int> t[105];
bool vis[105];
int s,len;
void dfs(int x,int d){
    vis[x]=true;
    if(len<d){
        s=x;
        len=d;
    }
    for(int i=0;i<t[x].size();i++){
        int o=t[x][i];
        if(vis[o]==false){
            dfs(o,d+1);
        }
    }
}
int main(){
    cin>>n;
    for(int i=1;i<n;i++){
        int x,y;
        cin>>x>>y;
        t[x].push_back(y);
        t[y].push_back(x);
    }
    memset(vis,0,sizeof(vis));
    dfs(1,0);
    memset(vis,0,sizeof(vis));
    dfs(s,0);
    cout<<len;
    return 0;
}
/*
提交时间:202311242117
解题时间:202311242045-202311242117
测试数据:
in1:
6
1 2
3 2
5 6
2 4
5 2
out1:
3

in2:
6
1 2
1 3
1 4
1 5
1 6
out2:
2

in3:
6
1 2
1 3
2 4
2 5
4 6
out3:
4

in4:
6
2 1
1 3
2 5
2 6
2 4
out4:
3

in5:
6
1 2
2 3
3 4
4 5
5 6
out5:
5
*/