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【题解】HackerRank - Coprime Paths

题目

https://www.hackerrank.com/challenges/coprime-paths/problem

题意

给一棵树,树上的每一个点都有一个至多有三个素因子的数,询问路径上互质点对个数。

题解

  • 树上莫队 + 容斥
  • 用的树上莫队的板子比较少见,其中 bug 变量的含义是如果不考虑就会翻转错一个点,好处就是不用求 LCA 并单独考虑。
  • 增加时加的是 cnt - 1,这样可以不用考虑自己与自己不互素,于是 1 的情况不用特判。
  • 容斥部分代码重用有些麻烦。

代码

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#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifdef zerol
#define dbg(args...) do { cout << "\033[32;1m" << #args<< " -> "; err(args); } while (0)
#else
#define dbg(...)
#endif
void err() { cout << "\033[39;0m" << endl; }
template<typename T, typename... Args>
void err(T a, Args... args) { cout << a << ' '; err(args...); }
// -----------------------------------------------------------------------------
const int maxn = 25E3 + 100;
const int maxp = 1E7 + 10;
int v[maxn], dep[maxn], fa[maxn], blk[maxn], in[maxn];
vector<int> G[maxn], p[maxn];
int B, ans[maxn], bug, Ans, L;
int cnt[maxp];
bool vis[maxn];

struct Q {
int u, v, idx;
bool operator < (const Q& b) const {
const Q& a = *this;
return blk[a.u] < blk[b.u] || (blk[a.u] == blk[b.u] && in[a.v] < in[b.v]);
}
};
vector<Q> query;

const int p_max = 1E5 + 100;
int prime[p_max], p_sz;
bool p_vis[p_max];
void get_prime() {
FOR (i, 2, p_max) {
if (!p_vis[i]) prime[p_sz++] = i;
FOR (j, 0, p_sz) {
if (prime[j] * i >= p_max) break;
p_vis[prime[j] * i] = 1;
if (i % prime[j] == 0) break;
}
}
}

void dfs(int u = 1, int d = 0) {
static int clk = 0, S[maxn], sz = 0, blk_cnt = 0;
dep[u] = d;
in[u] = clk++;
int btm = sz;
for (int v: G[u]) {
if (v == fa[u]) continue;
fa[v] = u;
dfs(v, d + 1);
if (sz - btm >= B) {
while (sz > btm) blk[S[--sz]] = blk_cnt;
++blk_cnt;
}
}
S[sz++] = u;
if (u == 1) while (sz) blk[S[--sz]] = blk_cnt - 1;
}

inline void up(int t, int sgn, int cnt_upd) {
if (cnt_upd) cnt[t] += cnt_upd;
else Ans += (cnt[t] - 1) * sgn;
}

void add(int k, int sgn, int cnt_upd = 0) {
FOR (i, 0, p[k].size())
up(p[k][i], sgn, cnt_upd);
FOR (i, 0, p[k].size())
FOR (j, i + 1, p[k].size())
up(p[k][i] * p[k][j], -sgn, cnt_upd);
if (p[k].size() == 3) up(p[k][0] * p[k][1] * p[k][2], sgn, cnt_upd);
}

void flip(int k) {
int sgn = vis[k] ? -1 : 1;
if (sgn == -1) add(k, -1);
add(k, 0, sgn);
if (sgn == 1) add(k, 1);
L += sgn;
vis[k] ^= 1;
}

void go(int& k) {
if (!bug) {
if (vis[k] && !vis[fa[k]]) bug = k;
if (!vis[k] && vis[fa[k]]) bug = fa[k];
}
flip(k);
k = fa[k];
}

void mv(int a, int b) {
bug = 0;
if (vis[b]) bug = b;
if (dep[a] < dep[b]) swap(a, b);
while (dep[a] > dep[b]) go(a);
while (a != b) go(a), go(b);
go(a); go(bug);
}

int main() {
#ifdef zerol
freopen("in", "r", stdin);
#endif
get_prime();
int n, q_sz;
cin >> n >> q_sz; B = int(sqrt(n));
FOR (i, 1, n + 1) {
scanf("%d", &v[i]);
int t = v[i];
FOR (j, 0, p_sz)
if (t % prime[j] == 0) {
p[i].push_back(prime[j]);
while (t % prime[j] == 0) t /= prime[j];
}
if (t != 1) p[i].push_back(t);
}
FOR (_, 1, n) {
static int u, v;
scanf("%d%d", &u, &v);
G[u].push_back(v); G[v].push_back(u);
}
dfs();
FOR (i, 1, n + 1) dbg(i, in[i], blk[i], p[i].size(), v[i], fa[i]);
FOR (i, 0, q_sz) {
static int u, v;
scanf("%d%d", &u, &v);
query.push_back({u, v, i});
}
sort(query.begin(), query.end());
int u = 1, v = 1; flip(1);
for (Q& q: query) {
mv(u, q.u); u = q.u;
mv(v, q.v); v = q.v;
dbg(q.idx, q.u, q.v, L, Ans);
ans[q.idx] = L * (L - 1) / 2 - Ans;
}
FOR (i, 0, q_sz) printf("%d\n", ans[i]);
}