[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[N], 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[N]; 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[N]; Node r[N]; 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[N], a[N]; 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; }