记录“得道者”的倍数,然后dfs一遍关系图,同时维护这一代的功力,遇到得道者就乘上倍数加到结果里即可 代码: #include<bits/stdc++.h>usingnamespacestd;constintmaxn =1e5+5; vector<int> G[maxn];intn, mul[maxn];doublez, r, sum;voiddfs(intu,doubleleft){if(mul[u]) sum += left * mul[u]...