题解

用bfs对每个点跑一遍最短路即可

#include <algorithm>
#include <bits/stdc++.h>
#include <cstring>
#include <queue>
using namespace std;

const int N = 2005;
const int INF = 1e9;
int n,m;
vector<int> g[N];
bool vis[N];
int d[N];
queue<int>q;
void bfs(int s){
    memset(vis , 0 , sizeof(vis));
    vis[s] = 1;
    q.push(s);
    d[s] = 0;
    while(!q.empty()){
        int t = q.front();
        q.pop();
        vis[t] = 1;
        for(int u : g[t]){
            if(vis[u] != 1){
                d[u] = d[t] + 1;
                vis[u] = 1;
                q.push(u);
            }
        }
    }
}
int minn = INF,ans;
int main(){
    cin >> n >> m;
    for(int i = 1; i <= m; i++){
        int u,v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    for(int i = 1; i <= n; i ++){
        bfs(i);
        int maxn = *max_element(d + 1, d + n + 1);
        if(minn > maxn){
            ans = i;
            minn = maxn;
        }
    }
    cout << ans;
    return 0;
}