[hdu4670 Cube number on a tree]点分治

题意：给一个N个带权节点的树，权值以给定的K个素数为因子，求路径上节点乘积为立方数的路径条数

思路：立方数的性质是每个因子的个数为3的倍数，那么每个因子只需要保存0-2三个状态即可，然后路径就可以转化为一个K位3进制数，点分治后，便可以用一个map来查询路径经过根的答案。代码与上一题（poj1741）类似：http://www.cnblogs.com/jklongint/p/4960052.html

#pragma comment(linker,"/STACK:10240000,10240000")
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <map>
#include <vector>
using namespace std;
#define X first
#define Y second
#define pb(x) push_back(x)
#define mp(x, y) make_pair(x, y)
#define all(a) (a).begin(), (a).end()
#define mset(a, x) memset(a, x, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(b))
#define cas() int T, cas = 0; cin >> T; while (T --)
template<typename T>bool umax(T&a, const T&b){return a<b?(a=b,true):false;}
template<typename T>bool umin(T&a, const T&b){return b<a?(a=b,true):false;}
typedef long long ll;
typedef pair<int, int> pii;

#ifndef ONLINE_JUDGE
#include "local.h"
#endif

const int N = 5e4 + 7;
const int M = N;
const int inf = 1e9 + 7;

namespace Edge {
int last
, to[M << 1], next[M << 1], cntE;
void init() {
cntE = 0;
memset(last, -1, sizeof(last));
}
void addEdge(int u, int v) {
to[cntE] = v;
next[cntE] = last[u];
last[u] = cntE ++;
}
}

int n, K;

struct Node {
char p[33];
char &operator[] (int x) {
return p[x];
}
ll getNum() {
ll ans = 0;
for (int i = 0; i < K; i ++) {
ans = ans * 3 + p[i];
}
return ans;
}
ll getComplement() {
ll ans = 0;
for (int i = 0; i < K; i ++) {
ans = ans * 3 + (p[i]? 3 - p[i] : 0);
}
return ans;
}
Node operator+ (Node &that) {
Node ans;
for (int i = 0; i < K; i ++) {
ans[i] = p[i] + that[i];
if (ans[i] >= 3) ans[i] -= 3;
}
return ans;
}
Node operator- (Node &that) {
Node ans;
for (int i = 0; i < K; i ++) {
ans[i] = p[i] - that[i];
if (ans[i] < 0) ans[i] += 3;
}
return ans;
}
};

namespace Center {
int root, siz, son
;
void init() {
siz = inf;
}
void getRoot(int cur, int fa, int total, bool used[]) {
son[cur] = 0;
int buf = 0;
for (int i = Edge::last[cur]; ~i; i = Edge::next[i]) {
int to = Edge::to[i];
if (to != fa && !used[to]) {
getRoot(to, cur, total, used);
son[cur] += son[to] + 1;
buf = max(buf, son[to] + 1);
}
}
buf = max(buf, total - son[cur] - 1);
if (buf < siz || buf == siz && cur < siz) {
siz = buf;
root = cur;
}
}
}

bool used
;
Node r
;

void getNode(int cur, int fa, Node sum, vector<Node> &vt, bool used[]) {
vt.pb(sum);
for (int i = Edge::last[cur]; ~i; i = Edge::next[i]) {
int to = Edge::to[i];
if (to != fa && !used[to]) getNode(to, cur, sum + r[to], vt, used);
}
}

ll getAns(vector<Node> &vt, Node &s) {
ll ans = 0;
map<ll, int> mp;
for (int i = 0; i < vt.size(); i ++) {
mp[vt[i].getNum()] ++;
ans += mp[(vt[i] - s).getComplement()];
}
return ans;
}

ll work(int cur) {
used[cur] = true;
vector<Node> total;
total.push_back(r[cur]);
ll ans = 0;
for (int i = Edge::last[cur]; ~i; i = Edge::next[i]) {
int to = Edge::to[i];
if (!used[to]) {
vector<Node> local;
getNode(to, cur, r[cur] + r[to], local, used);
ans -= getAns(local, r[cur]);
for (int j = 0; j < local.size(); j ++) {
total.push_back(local[j]);
}
Center::init();
Center::getRoot(to, cur, local.size(), used);
ans += work(Center::root);
}
}
return ans += getAns(total, r[cur]);
}

ll p
, a
;

int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
while (cin >> n >> K) {
Edge::init();
Center::init();
mset(r, 0);
mset(used, 0);
for (int i = 0; i < K; i ++) {
scanf("%I64d", p + i);
}
for (int i = 1; i <= n; i ++) {
scanf("%I64d", a + i);
for (int j = 0; j < K; j ++) {
ll cur = p[j];
while (a[i] % cur == 0) {
r[i][j] ++;
if (r[i][j] == 3) r[i][j] = 0;
cur *= p[j];
}
}
}
int u, v;
for (int i = 1; i < n; i ++) {
scanf("%d%d", &u, &v);
Edge::addEdge(u, v);
Edge::addEdge(v, u);
}
Center::getRoot(1, 0, n, used);
cout << work(Center::root) << endl;
}
return 0;
}