Codeforces Round #372 (Div. 2) E. Digit Tree （点分治）

题意

• 给一棵树，有n(n≤105)n(n≤105)个点。
• 边有权值wi(1≤wi≤9)wi(1≤wi≤9)。
• 求路径(u,v)(u,v)使得u→vu→v路径构成的大数模m为0，其中gcd（10，m）=1.
  

思路

• 点分治
• 根据欧拉定理
  xϕ(m)≡1(mod m),gcd(x,m)=1
  
  可以求出10的逆元。

#include<cmath>#include<cstdio>#include<cstring>#include<set>#include<map>#include<queue>#include<vector>#include<assert.h>#include<iostream>#include<algorithm>using namespace std;#define fi first#define se second#define pb push_back#define mp make_pair#define sz(x) ((int)(x).size())#define rep(i,l,r) for(int i=(l);i<(r);++i)#define setIO(x) freopen(x".in","r", stdin);freopen(x".out","w",stdout);typedef long long ll;typedef unsigned long long ul;typedef pair<int, int> pii;const ul BASE = 107;const ll LINF = 1e16 + 7;const int N = 1e5 + 7;const int INF = 1e9 + 7;const int MOD = 1000000007;const double PI = acos(-1.0);const double EPS = 1e-8;//-----------------head-----------------int n, m, q[N], f[N], msz[N], siz[N], pw[N], rpw[N];bool vis[N];ll cnt1, cnt2;map<int, int> R;vector<pii> e[N];ll kpow(ll a, ll b, ll mod) {ll r = 1;while (b > 0) {if (b & 1)r = r * a % mod;a = a * a % mod, b >>= 1;}return r;}int find(int u) {int t = 1;q[0] = u, f[u] = -1;rep(h, 0, t){u = q[h];msz[u] = 0, siz[u] = 1;rep(i, 0, sz(e[u])){int v = e[u][i].fi;if (vis[v] || v == f[u])continue;f[v] = u, q[t++] = v;}}for (int h = t - 1; h >= 0; --h) {u = q[h];msz[u] = max(msz[u], t - siz[u]);if (msz[u] * 2 <= t)return u;if (~f[u]) {siz[f[u]] += siz[u];msz[f[u]] = max(msz[f[u]], siz[u]);}}return 0;}void dfs1(int u, int w, int d, int p) {cnt1 += w == 0;cnt2 += R[1ll * (m - w) * rpw[d] % m];rep(i, 0, sz(e[u])){int v = e[u][i].fi;if (vis[v] || v == p)continue;dfs1(v, (1ll * 10 * w + e[u][i].se) % m, d + 1, u);}}void dfs2(int u, int w, int d, int p) {cnt1 += w == 0;++R[w];rep(i, 0, sz(e[u])){int v = e[u][i].fi;if (vis[v] || v == p)continue;dfs2(v, (1ll * e[u][i].se * pw[d] + w) % m, d + 1, u);}}void solve(int u) {u = find(u);R.clear();rep(i, 0, sz(e[u])){int v = e[u][i].fi;if (vis[v])continue;dfs1(v, e[u][i].se % m, 1, u);dfs2(v, e[u][i].se % m, 1, u);}reverse(e[u].begin(), e[u].end());R.clear();rep(i, 0, sz(e[u])){int v = e[u][i].fi;if (vis[v])continue;dfs1(v, e[u][i].se % m, 1, u);dfs2(v, e[u][i].se % m, 1, u);}vis[u] = true;rep(i, 0, sz(e[u]))if (!vis[e[u][i].fi])solve(e[u][i].fi);}int phi(int n) {int ret = n;for (int i = 2, I = sqrt(n + 0.5); i <= I; ++i)if (n % i == 0) {ret = ret / i * (i - 1);while (n % i == 0)n /= i;}if (n > 1)ret = ret / n * (n - 1);return ret;}void init(int n) {int inv = kpow(10, phi(m) - 1, m);pw[0] = rpw[0] = 1;rep(i, 1, n + 1){pw[i] = 1ll * 10 * pw[i - 1] % m;rpw[i] = 1ll * inv * rpw[i - 1] % m;}}int main() {scanf("%d%d", &n, &m);init(n);rep(i, 1, n){int u, v, w;scanf("%d%d%d", &u, &v, &w);e[u].pb(mp(v, w)), e[v].pb(mp(u, w));}memset(vis, false, sizeof(vis[0]) * n);cnt1 = cnt2 = 0;solve(0);printf("%I64d", cnt1 / 2 + cnt2);return 0;}