P2634 [国家集训队]聪聪可可

淀粉质

第二道点分治的题

关于点分治的一点理解：

所谓点分治，其实就是把要求的问题(一般与路径有关)划分成两种情况

1.路径经过rt(根节点)

2.路径在根节点的子树内

我们只需要处理情况1,因为情况2就是情况1的递归子问题

在这个过程中,要注意容斥原理的应用;

//--------------------------------------------------------------------------

关于此题:

w可预先%3(不会影响答案),注意乘法原理的应用;

 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 using namespace std;
 #define up(i,l,r) for(register int i = (l); i <= (r); ++i)
 #define dn(i,l,r) for(register int i = (l); i >= (r); --i)
 #define ll long long
 #define re register
 using namespace std;

 template <typename T> void in(T &x) {
     x = ; T f = ; char ch = getchar();
     while(!isdigit(ch)) {if(ch == '-') f = -; ch = getchar();}
     while( isdigit(ch)) {x =  * x + ch - ; ch = getchar();}
     x *= f;
 }

 template <typename T> void out(T x) {
     if(x < ) x = -x , putchar('-');
     if(x > ) out(x/);
     putchar(x% + );
 }
 //---------------------------------------------------------

 const int N = ;

 int n;
 ll ans = ;

 struct edge {
     int v,w,nxt;
 }e[N<<];int tot,head[N];

 void add(int u,int v,int w) {
     e[++tot].v = v;
     e[tot].w = w;
     e[tot].nxt = head[u];
     head[u] = tot;
 }

 //---------------------------------------------------------

 bool vis[N];
 int size[N],f[N];
 int Tsize,rt,dis[N];
 int t[];

 void get_rt(int u,int fa) {
     size[u] = ; f[u] = ;
     for(re int i = head[u]; i ;i = e[i].nxt) {
         int v = e[i].v; if(v == fa || vis[v]) continue;
         get_rt(v,u); size[u] += size[v];
         f[u] = max(f[u],size[v]);
     }
     f[u] = max(f[u],Tsize-size[u]);
     if(f[u] < f[rt]) rt = u;
 }

 void get_dis(int u,int fa) {
     ++t[dis[u]%];
     for(re int i = head[u]; i ;i = e[i].nxt) {
         int v = e[i].v; if(v == fa || vis[v]) continue;
         dis[v] = (dis[u]+e[i].w)%; get_dis(v,u);
     }
 }

 ll calc(int u) {
     t[] = t[] = t[] = ;
     get_dis(u,);
     return t[]*t[]+*t[]*t[];
 }

 void solve(int u) {
     vis[u] = ; dis[u] = ;
     ans += calc(u);
     for(re int i = head[u]; i ;i = e[i].nxt) {
         int v = e[i].v; if(vis[v]) continue;
         dis[v] = e[i].w;
         ans -= calc(v);
         Tsize = size[v],rt = ,f[] = n+;
         get_rt(v,); solve(rt);
     }
 }

 void init() {
     memset(head,,sizeof(head));
     memset(vis,,sizeof(vis));
     ans = ; Tsize = n,rt = ,f[] = n+;
 }

 ll Gcd(ll a,ll b) {return b ==  ? a:Gcd(b,a%b);}

 int main() {
     init();
     in(n); int x,y,w;
     up(i,,n-) {
         in(x); in(y); in(w); add(x,y,w%),add(y,x,w%);
         //技巧 mod 3
     }
     get_rt(,); solve(rt);
     ll gcd = Gcd(ans,1ll*n*n);
     out(ans/gcd); putchar('/'); out(1ll*n*n/gcd);
     return ;
 }