1. 利用哈希存储x1和x2的出现次数,y1和y2同理,如果x1出现一次,答案为x1,x2同理,y同理2. 前缀和+字符统计,根据count[26]判断是否可以变成回文字符串,如果出现count[k]为奇数的个数大于等于2则不是回文3. dfs,注意叶子节点的奇数权值和偶数权值个数的初始化代码pair<int, int> dfs(int u){st[u] = true;int odd = 0, even = 0;for (int i = h[u]; i != -1; i = ne[i]){int j = e[i];if (st[j]) continue;pair<int, int> s = dfs(j);if (w[i] % 2){ans += (s.second + 1) * odd;ans += (s.first) * odd;ans += s.first;even += s.first;odd += s.second + 1;}else{ans += (s.second + 1) * even;ans += (s.first) &amp; odd;even += s.second + 1;ans += s.second + 1;odd += s.first ;}}return {odd, even};}