题目链接：Atcoder 或者 洛谷
先考虑暴力，显然是枚举整棵树的路径，这个枚举复杂度显示是 \(O(n^2)\)，还不考虑计算 \(f(i,j)\)，考虑使用点分治优化枚举复杂度：\(O(n\log{n})\)。
接下来考虑如何计算每条路径的 \(f(i,j)\) ，注意到 \(f(i,j)\)：
当且仅当 \(a[i]=a[j]\) 时，答案加上 \(dis(i,j)\)，那么显然我们至少要维护一个与权值有关的桶信息。
接下来从点分治的角度思考：
- 对于拼链，我们选取其他子树的路径与当前子树的路径进行拼，考虑拼的情况：
我们枚举 \(a[当前子树的点]\)，显然当 \(a[当前子树的点]=a[其他子树的点]\)，就需要加上 \(dis(curr,other)\)，现在我们考虑其他子树所有相同值的点都一起计算，那么显然我们的公式为：
\[\sum_{a[curr]=a[other]} pre[curr]+pre[other]=cnt[other]\times pre[curr]+\sum pre[other] \]其中 \(pre\) 表示当前点到枚举的分治中心的距离，而 \(cnt[other]\) 表示其他子树满足 \(a[curr]=a[other]\) 的数量，那么显然是一个 \(权值\Rightarrow 数量\) 的桶。而显然 \(\sum pre[other]\) 这个东西，表示的是所有满足 \(a[curr]=a[other]\) 的前缀路径数量，那么这个玩意可以用一个类似的桶记录：\(cntVal[a[curr]]\) 就表示其他子树中值为 \(a[curr]\) 的前缀路径和，这样一来就解决了。
- 单独的链，我们把根节点视为一个前缀路径为 \(0\) 即 \(pre=0\) 的信息，这样一来就可以完美的归纳为上一个情况。
参照代码
#include <bits/stdc++.h>
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
#define isPbdsFile
#ifdef isPbdsFile
#include <bits/extc++.h>
#else
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
#endif
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
template <typename T>
int disc(T* a, int n)
{
return unique(a + 1, a + n + 1) - (a + 1);
}
template <typename T>
T lowBit(T x)
{
return x & -x;
}
template <typename T>
T Rand(T l, T r)
{
static mt19937 Rand(time(nullptr));
uniform_int_distribution<T> dis(l, r);
return dis(Rand);
}
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
return (a % b + b) % b;
}
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
a %= c;
T1 ans = 1;
for (; b; b >>= 1, (a *= a) %= c) if (b & 1) (ans *= a) %= c;
return modt(ans, c);
}
template <typename T>
void read(T& x)
{
x = 0;
T sign = 1;
char ch = getchar();
while (!isdigit(ch))
{
if (ch == '-') sign = -1;
ch = getchar();
}
while (isdigit(ch))
{
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
x *= sign;
}
template <typename T, typename... U>
void read(T& x, U&... y)
{
read(x);
read(y...);
}
template <typename T>
void write(T x)
{
if (typeid(x) == typeid(char)) return;
if (x < 0) x = -x, putchar('-');
if (x > 9) write(x / 10);
putchar(x % 10 ^ 48);
}
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
write(x), putchar(c);
write(c, y...);
}
template <typename T11, typename T22, typename T33>
struct T3
{
T11 one;
T22 tow;
T33 three;
bool operator<(const T3 other) const
{
if (one == other.one)
{
if (tow == other.tow) return three < other.three;
return tow < other.tow;
}
return one < other.one;
}
T3()
{
one = tow = three = 0;
}
T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
{
}
};
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
if (x < y) x = y;
}
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
if (x > y) x = y;
}
constexpr int N = 2e5 + 10;
int siz[N], root, sumSize, maxSon;
bool del[N];
vector<int> child[N];
ll cntVal[N];
ll cnt[N];
int n;
int a[N];
ll res;
inline void dfs(const int curr, const int fa)
{
siz[curr] = 1;
for (const int nxt : child[curr]) if (nxt != fa and !del[nxt]) dfs(nxt, curr), siz[curr] += siz[nxt];
}
inline void makeRoot(const int curr, const int fa)
{
siz[curr] = 1;
int currSize = 0;
for (const int nxt : child[curr])
{
if (nxt == fa or del[nxt]) continue;
makeRoot(nxt, curr);
siz[curr] += siz[nxt];
uMax(currSize, siz[nxt]);
}
uMax(currSize, sumSize - siz[curr]);
if (currSize < maxSon) maxSon = currSize, root = curr;
}
//权值,前缀路径长
pll dis[N];
int tot;
ll pre[N];
inline void getDis(const int curr, const int fa)
{
pre[curr] = pre[fa] + 1; //算出前缀路径长
dis[++tot] = pii(a[curr], pre[curr]); //(值,前缀路径)
for (const int nxt : child[curr])
{
if (nxt == fa or del[nxt]) continue;
getDis(nxt, curr);
}
}
inline void cale(const int curr)
{
vector<pll> add;
cnt[a[curr]]++; //这个根的值的数量
pre[curr] = 0; //重置当前点为根的前缀路径长
for (const auto nxt : child[curr])
{
if (del[nxt]) continue;
tot = 0;
getDis(nxt, curr);
//枚举a[other]
forn(i, 1, tot)
{
const auto [v,len] = dis[i];
add.emplace_back(v, len); //用于删除桶信息的
res += cnt[v] * len + cntVal[v]; //计算公式
}
//当前子树信息加入到其他子树中
forn(i, 1, tot)
{
const auto [v,len] = dis[i];
cntVal[v] += len;
cnt[v]++;
}
}
//撤销信息
for (const auto [v,len] : add) cnt[v]--, cntVal[v] -= len;
cnt[a[curr]]--;
}
inline void div(const int curr)
{
cale(curr);
del[curr] = true;
for (const int& nxt : child[curr])
{
if (del[nxt]) continue;
maxSon = sumSize = siz[nxt];
makeRoot(nxt, 0);
dfs(nxt, 0);
div(root);
}
}
inline void solve()
{
cin >> n;
forn(i, 1, n-1)
{
int u, v;
cin >> u >> v;
child[u].push_back(v);
child[v].push_back(u);
}
forn(i, 1, n) cin >> a[i];
sumSize = maxSon = n;
makeRoot(1, 0);
dfs(root, 0);
div(root);
cout << res;
}
signed int main()
{
// MyFile
Spider
//------------------------------------------------------
// clock_t start = clock();
int test = 1;
// read(test);
// cin >> test;
forn(i, 1, test) solve();
// while (cin >> n, n)solve();
// while (cin >> test)solve();
// clock_t end = clock();
// cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}