CodeForces 109C 树形DP Lucky Tree

赶脚官方题解写得挺清楚的说，=_=

注意数据范围用long long，否则会溢出。

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 using namespace std;

 const int maxn =  + ;

 int n;
 vector<int> G[maxn], C[maxn];

 inline int lucky(int x)
 {
     while(x)
     {
         if(x %  !=  && x %  != ) return ;
         x /= ;
     }
     return ;
 }

 long long d[maxn], sz[maxn], f[maxn];

 void dfs(int u, int fa)
 {
     sz[u] = ;
     f[u] = ;
     for(int i = ; i < G[u].size(); i++)
     {
         int v = G[u][i];
         if(v == fa) continue;
         dfs(v, u);
         sz[u] += sz[v];
         if(C[u][i]) f[u] += sz[v];
         else f[u] += f[v];
     }
 }

 void dfs2(int u, int fa)
 {
     for(int i = ; i < G[u].size(); i++)
     {
         int v = G[u][i];
         if(v == fa) continue;
         if(C[u][i]) d[v] = sz[] - sz[v];
         else d[v] = d[u] + f[u] - f[v];
         dfs2(v, u);
     }
 }

 int main()
 {
     //freopen("in.txt", "r", stdin);

     scanf("%d", &n);
     for(int i = ; i < n; i++)
     {
         int u, v, d; scanf("%d%d%d", &u, &v, &d);
         int t = lucky(d);
         G[u].push_back(v); C[u].push_back(t);
         G[v].push_back(u); C[v].push_back(t);
     }

     dfs(, );
     dfs2(, );

     long long ans = ;
     for(int i = ; i <= n; i++) ans += f[i] * (f[i] - ) + d[i] * (d[i] - ) +  * f[i] * d[i];
     printf("%I64d\n", ans);

     return ;
 }

代码君