树的重心：
    假设有unordered_map<int,vector<int>> m 表示与结点关联的结点
    void dfs(int v,int fa){
        son[v] = 0;
        int d = G[v].size();
        int prebalance = 0;
        for(int i=0;i<d;i++){
            int k = G[v][i];
            if(k==fa) continue;
            son[v] += son[u]+1;
            prebalance = max(prebalance,son[u]+1);
        }
        prebalance = max(prebalance,n-son[v]-1);
        if(prebalance<balance || balance == prebalance && v<ans){
            balance = prebalance;
            ans = v;
        }
    }

树的最长路径：
    int lastorder(int pre,int cur){
        int first = 0, second = 0;
        for(int i = 0;i<G[cur].size();i++){
            if(G[cur][i]==pre) continue;
            int temp = lastorder(cur,G[cur][i]);
            if(temp>first){
                second = first;
                first = temp;
            }else if(temp>second){
                second = temp;
            }
        }
        maxdistance = max(maxdistance,first+second);
        return first+1;
    }