分析：題意不難理解，看上去就是不太難的題，然後WA了一天qaq

用bfs和並查集應該都能做，然後我沒壓縮好，一會超時一會超記憶體。最後用模擬stack儲存輩分數降低了遞迴的時間複雜度。

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<vector>
 4 #include<map>
 5 #include<stack>
 6 using namespace std;
 7 const int N = 1e5+10;
 8 
 9 vector<int>ans;
 
10 map<int,int>mp;
11 int cnt=0;
12 int fa[N];
13 int gen[N];
14 stack<int>q;
15 void fstack(int n)
16 {
17     while(!q.empty())
18         q.pop();
19     for(int i = 1; i < n; i ++)
20     {
21         if(fa[i]==0)
22             gen[i]=1;
23         else
24         {
25             q.push(i);
 
26             while(gen[fa[q.top()]]!=0)
27             {
28                 gen[q.top()] = gen[fa[q.top()]]+1;
29                 q.pop();
30                 if(q.empty())
31                     break;
32             }
33         }
34     }
35 }
36 int main()
37 {
38     int n,m,a,flag;
39     double
 
 z,r,bei[N]= {0},sum=0;
40     scanf("%d%lf%lf",&n,&z,&r);
41     fa[0] = 0;
42     for(int  i = 0; i < n; i ++)
43     {
44         scanf("%d",&m);
45         if(m==0)
46         {
47             scanf("%d",&a);
48             ans.push_back(i);
49             mp[i]=a;
50         }
51         else
52         {
53             for(int j = 0; j < m; j ++)
54             {
55                 scanf("%d",&a);
56                 fa[a]=i;
57             }
58         }
59     }
60     fstack(n);
61     bei[0] = z;
62     double k = (100-r)/(double)100;
63     for(int i = 1; i < n; i ++)
64     {
65         bei[i] = bei[i-1]*k;
66     }
67     vector<int>::iterator it;
68     for(it = ans.begin(); it!=ans.end(); it++)
69     {
70         flag = *it;
71         sum += bei[gen[flag]]*mp[flag];
72     }
73     int as = sum;
74     printf("%d\n",as);
75     return 0;
76 }