[leetcode] 2049 统计最高分的节点数目 | dfs二叉树

Posted 2022-03-15 PushyTao

题目链接

记录父亲节点的深度优先遍历不经常写，然后把给出的数据改成记录子节点，然后对根进行$dfs$，记录以当前节点为根的结点的数量，然后枚举删除某个节点的情况下的分数是多少
需要讨论当前节点是否为根

然后统计最大值并记录个数

Code：

class Solution 
public:
    int cnt[100000 + 1];
    vector<int> son[100000 + 1];
    int countHighestScoreNodes(vector<int> &parents) 
        const int siz = parents.size();
        if (siz == 2) return 2;
        memset(cnt, 0, sizeof cnt);
        int root;
        for (int i = 0; i < siz; i++) 
            int fa = parents[i];
            if (fa == -1) 
                root = i;
                continue;
            
            son[fa].push_back(i);
        
        cnt[root] = dfs(root);
        long long mx = 0, ret = 0;
        for (int i = 0; i < siz; i++) 
            if (root == i) 
                long long tmx = 1;
                for (int x: son[i]) tmx *= cnt[x];
                if (tmx < mx) continue;
                else if (tmx == mx) ret++;
                else 
                    mx = tmx;
                    ret = 1;
                
             else 
                long long tmx = 1;
                for (int x: son[i]) tmx *= cnt[x];
                tmx *= siz - cnt[i];
                if (tmx < mx) continue;
                else if (tmx == mx) ret++;
                else 
                    mx = tmx;
                    ret = 1;
                
            
            // cout << i << "   " << mx << endl;
        
        return ret;
    

    int dfs(int root) 
        int sum = 0;
        for (int i = 0; i < son[root].size(); i++) 
            int u = son[root][i];
            cnt[u] = dfs(u);
            sum += cnt[u];
        
        return sum + 1;
    
;