D树

时间限制：10000/5000 MS（Java / Others）内存限制：102400/102400 K（Java / Others）
总共提交5400个已接受的提交1144

点分治

这种树上找路径问题最容易想到的就是点分治

点治的思想其实很简单，分别以每个点为根，找出所有经过根的路径更新答案

由于路径是一个二维的量，直接枚举是O(n^2)，而点分治通过固定一个根而使问题简化为一维O(n)

而由于树的性质，只要我们每次求出重心就可以保证最多只有logn层

总的复杂度就成了O(每一层操作复杂度 * logn)一般都是O(nlogn)或O(nlog^2n)

然而我点分治还是很生疏【我还是太弱了】

对于这道题，我们需要找到两条路径权值乘积取模为K

对于x * y ≡ K (mod P)，可以化为x ≡ K/y (mod P)

所以我们只需开一个hash表存x的值，对于每个y，用K乘上y的逆元查表更新答案就好了

要注意的细节就是根节点也要算上，而且查找与更新的路径只能有一个经过根，也就是算y时不算上，而算x存表时算上根

继续练习吧

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = head[u]; k != -1; k = edge[k].next)
using namespace std;
const int maxn = 100005,maxm = 200005,INF = 1000000000;
const LL P = 1000003;
inline int RD(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();}return out * flag;
}
int N,K,V[maxn],Siz[maxn],F[maxn],vis[maxn],rt,sum,ansx,ansy;
LL Hash[P],tmp[maxn],d[maxn],id[maxn],inv[P],cnt = 0;
int head[maxn],nedge = 0;
struct EDGE{int to,next;}edge[maxm];
inline void build(int u,int v){edge[nedge] = (EDGE){v,head[u]}; head[u] = nedge++;edge[nedge] = (EDGE){u,head[v]}; head[v] = nedge++;
}
void getRT(int u,int fa){int to; Siz[u] = 1; F[u] = 0;Redge(u) if (!vis[to = edge[k].to] && to != fa){getRT(to,u);Siz[u] += Siz[to];F[u] = max(F[u],Siz[to]);}F[u] = max(F[u],sum - Siz[u]);if (F[u] < F[rt]) rt = u;
}
inline void query(int x,int u){x = 1ll * inv[x] * K % P;int v = Hash[x];if (!v) return;if (v < u) swap(u,v);if (u < ansx || (u == ansx && v < ansy))ansx = u,ansy = v;
}
void dfs(int u,int fa){tmp[++cnt] = d[u]; id[cnt] = u; int to;Redge(u) if (!vis[to = edge[k].to] && to != fa){d[to] = 1ll * V[to] * d[u] % P;dfs(to,u);}
}
void solve(int u){int to; vis[u] = true; Hash[V[u]] = u;Redge(u) if (!vis[to = edge[k].to]){cnt = 0; d[to] = V[to];dfs(to,u);REP(i,cnt) query(tmp[i],id[i]);cnt = 0; d[to] = 1ll * V[to] * V[u] % P;dfs(to,u);REP(i,cnt) if (!Hash[tmp[i]] || Hash[tmp[i]] > id[i]) Hash[tmp[i]] = id[i];}Hash[V[u]] = 0;Redge(u) if (!vis[to = edge[k].to]){cnt = 0; d[to] = 1ll * V[to] * V[u] % P;dfs(to,u);REP(i,cnt) Hash[tmp[i]] = 0;}Redge(u) if (!vis[to = edge[k].to]){sum = Siz[to]; F[rt = 0] = INF;getRT(to,rt);solve(rt);}
}
void init(){memset(vis,0,sizeof(vis));memset(head,-1,sizeof(head)); nedge = 0; ansx = ansy = INF;REP(i,N) V[i] = RD() % P;REP(i,N - 1) build(RD(),RD());
}
void INIT(){inv[1] = 1;for (int i = 2; i < P; i++){inv[i] = ((P - P / i) * inv[P % i] % P + P) % P;}
}
int main(){INIT();while (~scanf("%d%d",&N,&K)){init();F[rt = 0] = INF; sum = N;getRT(1,rt);solve(rt);if (ansx == INF) printf("No solution\n");else printf("%d %d\n",ansx,ansy);}return 0;
}