CF387D.md

枚举每个点为中心点，然后剩下的点需要用圈覆盖，找最大匹配即可

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 555;
int n,m,match[MAXN];
bool mp[MAXN][MAXN],vis[MAXN];
vector<int> G[MAXN];
bool dfs(int u, int c){
    vis[u] = true;
    for(int v : G[u]){
        if(v==c) continue;
        if(match[v]==-1 or (!vis[match[v]] and dfs(match[v],c))){
            match[v] = u;
            return true;
        }
    }
    return false;
}
int hungary(int c){
    memset(match,255,sizeof(match));
    int mt = 0;
    for(int i = 1; i <= n; i++){
        if(i==c) continue;
        memset(vis,0,sizeof(vis));
        if(dfs(i,c)) mt++;
    }
    return mt;
}
int main(){
    ____();
    cin >> n >> m;
    for(int i = 1; i <= m; i++){
        int u, v; cin >> u >> v;
        G[u].push_back(v);
        mp[u][v] = true;
    }
    int ret = MAXN * MAXN * MAXN;
    for(int i = 1; i <= n; i++){
        int centeredge = 0;
        for(int j = 1; j <= n; j++){
            if(i==j and mp[i][j]) centeredge++;
            else if(i!=j){
                if(mp[i][j]) centeredge++;
                if(mp[j][i]) centeredge++;
            }
        }
        int mt = hungary(i);
        ret = min(ret,3*n+m-2-2*centeredge-2*mt);
    }
    cout << ret << endl;
    return 0;
}