神仙题。。orz yyb
考虑点分治，那么每次我们只需要统计以当前点为\(lca\)的点对之间的贡献以及\(lca\)到所有点的贡献。
一个很神仙的思路是，对于任意两个点对的路径上的颜色，我们只统计里根最近的那个点的贡献。
有了这个思路我们就可以瞎搞了，具体的细节很繁琐，但是大概思路是事实维护每个点的子树中的点会产生的贡献。比如某个点的颜色在它到根的路径上第一次出现，那么它子树中的所有点\(siz[x]\)，都会对外面的点产生贡献。
统计子树的时候只需要先消除掉子树的影响，然后dfs的时候考虑一下新加的颜色的贡献。。
复杂度\(o(n \log n)\)
我要评论#include<bits/stdc++.h>
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
#define pb push_back
using namespace std;
const int maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
const double eps = 1e-9;
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
template <typename a, typename b> inline ll fp(a a, b p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}
template <typename a> a inv(a x) {return fp(x, mod - 2);}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int n, c[maxn], cnt[maxn], vis[maxn], siz[maxn], lim, mx[maxn], root;
ll ans[maxn], num[maxn], sum;
vector<int> v[maxn];
void findroot(int x, int fa) {
siz[x] = 1; mx[x] = 1;
for(auto &to : v[x]) {
if(to == fa || vis[to]) continue;
findroot(to, x);
siz[x] += siz[to];
chmax(mx[x], siz[to]);
}
chmax(mx[x], lim - siz[x]);
if(mx[x] < mx[root])
root = x;
}
void dfs(int x, int fa, int opt) {
cnt[c[x]]++;
if(cnt[c[x]] == 1) sum += siz[x] * opt, num[c[x]] += siz[x] * opt;
for(auto &to : v[x])
if(to != fa && !vis[to]) dfs(to, x, opt);
cnt[c[x]]--;
}
void calc(int x, int fa) {
cnt[c[x]]++;
if(cnt[c[x]] == 1) sum += lim - num[c[x]];
ans[x] += sum;
for(auto &to : v[x]) {
if(to == fa || vis[to]) continue;
calc(to, x);
}
cnt[c[x]]--;
if(cnt[c[x]] == 0) sum -= lim - num[c[x]];
}
void divide(int x) {
if(vis[x]) return ; vis[x] = 1;
sum = 0; findroot(x, 0);
dfs(x, 0, 1); ans[x] += sum;
for(auto &to : v[x]) {
if(vis[to]) continue;
num[c[x]] -= siz[to]; sum -= siz[to]; lim -= siz[to];
cnt[c[x]] = 1; dfs(to, x, -1); cnt[c[x]] = 0;
calc(to, x);
cnt[c[x]] = 1; dfs(to, x, 1); cnt[c[x]] = 0;
num[c[x]] += siz[to]; sum += siz[to]; lim += siz[to];
}
dfs(x, 0, -1);
for(auto &to : v[x])
if(!vis[to]) {
root = 0, lim = siz[to], findroot(to, x);
divide(root);
}
}
signed main() {
//freopen("a.in", "r", stdin);freopen("b.out", "w", stdout);
n = read(); mx[0] = 1e9;
for(int i = 1; i <= n; i++) c[i] = read();
for(int i = 1; i < n ; i++) {
int x = read(), y = read();
v[x].pb(y); v[y].pb(x);
}
lim = n; root = 0; findroot(1, 0);
divide(root);
for(int i = 1; i <= n; i++) cout << ans[i] << '\n';
return 0;
}
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C语言字符数组应用示例1：编写一个程序，将两个字符串连接起来，不用strcat函数。
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