P2634 [国家集训队]聪聪可可

题意：

一颗n个点的树，问其中两点之间的边上数的和加起来是3的倍数的点对有多少个？
输出这样的点对所占比例

题解：

因为是求三的倍数，我们num来记录%3=0，1，2的数量，对于u得到的d[]，累加num[3-d[]]的数量。
其实就是常规的点分治模板

代码：

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{x= 0;char c= getchar();bool flag= 0;while (c < '0' || c > '9')flag|= (c == '-'), c= getchar();while (c >= '0' && c <= '9')x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();if (flag)x= -x;read(Ar...);
}
template <typename T> inline void write(T x)
{if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn= 2e5 + 9;
vector<PII> vec[maxn];
int vis[maxn], d[maxn], sz[maxn];
int rt, cnt;
int num[4];
void dfs_rt(int u, int fa, int tot)
{sz[u]= 1;int v, w, n= 0;for (auto it : vec[u]) {v= it.first;if (v == fa || vis[v])continue;dfs_rt(v, u, tot);sz[u]+= sz[v];n= max(n, sz[v]);}n= max(n, tot - sz[u]);if (2 * n <= tot)rt= u;
}
void dfs_dis(int u, int fa, int dis)
{d[++cnt]= dis;for (auto it : vec[u]) {int v= it.first;int w= it.second;if (v == fa || vis[v])continue;dfs_dis(v, u, dis + w);}
}
int work(int u, int fa, int tot)
{dfs_rt(u, fa, tot);u= rt;vis[u]= 1;ll ans= 0;for (auto it : vec[u]) {int v= it.first;int w= it.second;if (vis[v])continue;cnt= 0;dfs_dis(v, u, w);sz[v]= cnt;for (int i= 1; i <= cnt; i++) {ans+= num[(3 - d[i] % 3) % 3];if (d[i] % 3 == 0)ans++;}for (int i= 1; i <= cnt; i++) {num[d[i] % 3]++;}}num[0]= num[1]= num[2]= 0;for (auto it : vec[u]) {int v= it.first;if (vis[v])continue;ans+= work(v, u, sz[v]);}return ans;
}
int main()
{//rd_test();int n;read(n);for (int i= 1; i < n; i++) {int u, v, w;read(u, v, w);vec[u].push_back({v, w});vec[v].push_back({u, w});}ll a= 2ll * work(1, 0, n) + n;ll b= 1ll * n * n;ll g= __gcd(a, b);printf("%lld/%lld\n", a / g, b / g);return 0;//Time_test();
}