Ｐroblem
给一颗树，边有边权。求经过的边权积为完全平方数的路径条数。
\(n<1e5,a_i<1e8\)
Ｓolution
emm
可以看出要用点分治。
可是桶不好开。
所以要考虑转换。
我们可以将任意一个数X这样表示。
\(X = p_1^{a_1}\* p_2^{a_2} \* p_3^{a_3}\) ......
\(p\) 为质数。
然后我们给所有\(p_i\) 赋一个随机值\(h(p_i)\)。
定义 \(F(X)\) 为所有 \(a_i为奇数的h(p_i)的异或和\)
所以，我们可以得到。
当且仅当 \(F(x) = 0\) 时，\(x\) 为完全平方数。
所以我们可以把 \(A,B\) 两数的积用 \(F(A)\oplus F(B)\) 表示，因为我们仅需要知道一个数是否为完全平方数。
然后这题就完了。
好吧，还有一些细节。
比如 \(h(p_i)\) 的随机值要是 \(long~long\) 范围的。这样才能保证正确性。
还有，因为 \(a_i<1e8\) 不能用线性筛，所以我们要筛出 \(1e4\) 以内的质数，然后对所有数用 \(\frac {\sqrt a_i}{\ln n}\) 的复杂度算出 \(F(a_i)\) 。
还有一点很重要，桶不能用 \(map\) ，只能用 \(hash\) 。
先上一份自己打的，但是被 \(hack\) 数据卡T了的代码：
#pragma GCC optimize(2)
#pragma G++ optimize(2)
#pragma GCC optimize(3)
#pragma G++ optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define fst first
#define snd second
#define SZ(u) ((int) (u).size())
#define ALL(u) (u).begin(), (u).end()
template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline T read()
{
register T sum(0), fg(1);
register char ch(getchar());
for(; !isdigit(ch); ch = getchar()) if(ch == '-') fg = -1;
for(; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) - 48 + ch;
return sum * fg;
}
typedef long long LL;
typedef pair<int, LL> pii;
const int Maxn = 1e5 + 10, Maxv = 1e4 + 10, mod = 2333333;
vector <pii> Map[Maxn];
unordered_map <LL, LL> mp,book;
int pre[Maxn], top, Top, siz[Maxn], n;
bool vis[Maxn],bk[Maxn + 10];
long long p[Maxn] ,del[Maxn], pre_rand[Maxn + 10], stk[Maxn], ans;
LL Rand(){
int res = rand(),res2 = rand();
LL tmp = 1ll * res * res2;
while(tmp <= 0) {
if(res > res2) swap(res,res2);
res = rand();
tmp = 1ll * res * res2;
}
return tmp;
}
vector<pii> h[mod + 10];
int ask(LL number){
int now = number % mod;
for (int i = h[now].size() - 1;i >= 0; --i){
if(h[now][i].snd == number)
return h[now][i].fst;
}
return 0;
}
void add(LL number, int d){
int now = number % mod;
bool fl = 1;
for (int i = h[now].size() - 1;i >= 0; --i){
if(h[now][i].snd == number)
h[now][i].fst += d,fl = 0;
}
if(fl)h[now].push_back((pii){d,number});
}
void init(){
srand(19260817);
for (int i = 2;i <= Maxv; ++i){
if(!bk[i]){
pre[++Top] = i;
pre_rand[Top] = Rand();
for (int j = i << 1; j <= Maxv; j += i) bk[j] = 1;
}
}
}
void div_siz(int now, int pa){
siz[now] = 1;
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(vis[nxt] || nxt == pa)continue;
div_siz(nxt, now);
siz[now] += siz[nxt];
}
}
int div_rt(int now,int pa, int tot_siz){
bool fg = 1;
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(vis[nxt] || nxt == pa)continue;
int p = div_rt(nxt,now,tot_siz);
if(p) return p;
if((siz[nxt] << 1) > tot_siz) fg = 0;
}
if((tot_siz - siz[now] << 1) > tot_siz) fg = 0;
if(fg) return now;
return 0;
}
LL get_p(int number){
LL res = 0;
for (int i = 1; i <= Top && pre[i] <= number; ++i){
while(number % pre[i] == 0){
res ^= pre_rand[i];
number /= pre[i];
}
}
if (number > 1){
if (!mp[number]){
mp[number] = Rand();
}
res ^= mp[number];
}
return res;
}
void get_dis(int now, int pa){
stk[++top] = p[now];
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(vis[nxt] || nxt == pa)continue;
p[nxt] = p[now] ^ Map[now][i].snd;
get_dis(nxt, now);
}
}
void calc(int now){
int tot = 0;
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(vis[nxt]) continue;
top = 0;
p[nxt] = Map[now][i].snd;
get_dis(nxt, now);
for (int j = 1; j <= top; ++j){
ans += ask(stk[j]);
}
for (int j = 1; j <= top; ++j)
add(stk[j], 1), del[++tot] = stk[j];
}
for (int i = 1; i <= tot; ++i)
add(del[i], -1);
}
void div_solve(int now){
div_siz(now, 0), now = div_rt(now, 0, siz[now]), calc(now);
vis[now] = 1;
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(!vis[nxt]) div_solve(nxt);
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("a.in", "r", stdin);
freopen("a.out", "w", stdout);
#endif
n = read<int>();
init();
for (int i = 1; i < n; ++i){
int x = read<int>(), y = read<int>(), w = read<int>();
LL w0 = get_p(w);
Map[x].push_back((pii){y, w0});
Map[y].push_back((pii){x, w0});
}
add(0,1);
div_solve(1);
printf("%lld\n",2ll * ans);
return 0;
}
这份代码，卡了半年常，就是卡不过。
后来把手写哈希换成hash_table就过了。
感谢fatesky大佬的帮助！！！
#include <bits/stdc++.h>
using namespace std;
#define fst first
#define snd second
#define SZ(u) ((int) (u).size())
#define ALL(u) (u).begin(), (u).end()
template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline T read()
{
register T sum(0), fg(1);
register char ch(getchar());
for(; !isdigit(ch); ch = getchar()) if(ch == '-') fg = -1;
for(; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) - 48 + ch;
return sum * fg;
}
typedef long long LL;
typedef pair<int, LL> pii;
const int Maxn = 1e5 + 10, Maxv = 1e4 + 10, mod = 2333333;
vector <pii> Map[Maxn];
map <LL, LL> mp,book;
int pre[Maxn], top, Top, siz[Maxn], n;
bool vis[Maxn],bk[Maxn + 10];
long long p[Maxn],del[Maxn], pre_rand[Maxn + 10], stk[Maxn], ans;
LL Rand(){
int res = rand(),res2 = rand();
LL tmp = 1ll * res * res2;
while(tmp <= 0) {
if(res > res2) swap(res,res2);
res = rand();
tmp = 1ll * res * res2;
}
return tmp;
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
namespace DIV
{
cc_hash_table<LL, int> H;
void init(){
srand(19260817);
for (int i = 2;i <= Maxv; ++i){
if(!bk[i]){
pre[++Top] = i;
pre_rand[Top] = Rand();
for (int j = i << 1; j <= Maxv; j += i) bk[j] = 1;
}
}
}
void div_siz(int now, int pa){
siz[now] = 1;
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(vis[nxt] || nxt == pa)continue;
div_siz(nxt, now);
siz[now] += siz[nxt];
}
}
int div_rt(int now,int pa, int tot_siz){
#define p zjmsb
bool fg = 1;
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(vis[nxt] || nxt == pa)continue;
int p = div_rt(nxt,now,tot_siz);
if(p) return p;
if((siz[nxt] << 1) > tot_siz) fg = 0;
}
if((tot_siz - siz[now]) * 2 > tot_siz) fg = 0;
if(fg) return now;
return 0;
#undef p
}
LL get_p(int number){
LL res = 0;
for (int i = 1; i <= Top && pre[i] <= number; ++i){
while(number % pre[i] == 0){
res ^= pre_rand[i];
number /= pre[i];
}
}
if (number > 1){
if (!mp[number]) mp[number] = Rand();
res ^= mp[number];
}
return res;
}
void get_dis(int now, int pa){
stk[++top] = p[now];
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(vis[nxt] || nxt == pa)continue;
p[nxt] = p[now] ^ Map[now][i].snd;
get_dis(nxt, now);
}
}
void calc(int now){
H.clear();
++H[0];
int tot = 0;
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(vis[nxt]) continue;
top = 0;
p[nxt] = Map[now][i].snd;
get_dis(nxt, now);
for (int j = 1; j <= top; ++j) if(H.find(stk[j]) != H.end()) ans += H[stk[j]];
for (int j = 1; j <= top; ++j) ++H[stk[j]];
}
}
void div_solve(int now){
div_siz(now, 0), now = div_rt(now, 0, siz[now]), calc(now);
vis[now] = 1;
for (int i = Map[now].size() - 1; i >= 0; --i){
int nxt = Map[now][i].fst;
if(!vis[nxt]) div_solve(nxt);
}
}
}
using namespace DIV;
int main()
{
#ifndef ONLINE_JUDGE
freopen("zjm.in", "r", stdin);
freopen("zjm.out", "w", stdout);
#endif
n = read<int>();
init();
for (int i = 1; i < n; ++i){
int x = read<int>(), y = read<int>(), w = read<int>();
LL w0 = get_p(w);
Map[x].emplace_back(y, w0);
Map[y].emplace_back(x, w0);
}
div_solve(1);
printf("%lld\n",2ll * ans);
return 0;
}