HDU - 4812,点分治

D Tree

https://vjudge.net/problem/48318/origin
There is a skyscraping tree standing on the playground of Nanjing University of Science and Technology. On each branch of the tree is an integer (The tree can be treated as a connected graph with N vertices, while each branch can be treated as a vertex). Today the students under the tree are considering a problem: Can we find such a chain on the tree so that the multiplication of all integers on the chain (mod 10 6 + 3) equals to K?
Can you help them in solving this problem?

题意：给你一颗树,每个节点有一个权值val，问是否存在一个点对（a,b）使得路径上所有节点的权值的乘积等于k，如果存在，输出字典序最小的（a,b）
思路：点分治，在处理某个重心时，先算出它的一颗子树上的所有节点到根节点的距离的乘积，然后寻找一个x使得 $dis[i]*x = k*val[u]$，其中dis[i]为某一个乘积，需要找到另一颗子树上的一条乘积为x的路径，使得乘积为k*val[u](因为根节点乘了两遍，即 $x = k*val[u]*inv[dis[i]]$，一开始把所有逆元预处理出来，然后查找是否在其他子树出现过x即可，出现过就对点对进行更新

#include<bits/stdc++.h>
#define MAXN 100010
#define INF 0x3f3f3f3f
using namespace std;
const int MOD = 1e6+3;
int inv[MOD+5];
inline void getInv()
{
    inv[1] = 1;
    for(int i = 2;i < MOD;++i)
        inv[i] = 1ll*(MOD-MOD/i)*inv[MOD%i]%MOD;
}
int head[MAXN],tot;
struct edge
{
    int v,nxt;
}edg[MAXN << 1];
inline void addedg(int u,int v)
{
    edg[tot].v = v;
    edg[tot].nxt = head[u];
    head[u] = tot++;
}
int n,val[MAXN],k;
int mx,root,Size,sz[MAXN];
bool vis[MAXN];
inline void getroot(int u,int f)
{
    sz[u]  =1;
    int v,mson = 0;
    for(int i = head[u];i != -1;i = edg[i].nxt)
    {
        v = edg[i].v;
        if(v == f || vis[v])
            continue;
        getroot(v,u);
        sz[u] += sz[v];
        mson = max(mson,sz[v]);
    }
    mson = max(mson,Size-sz[u]);
    if(mson < mx)
        mx = mson,root = u;
}
int ans1,ans2;
int dis[MAXN],id[MAXN],cnt;
int visdis[MOD+5],visid[MOD+5],cc,cdis[MAXN];
inline void getdis(int u,int f,int d)
{
    int w = 1ll * d * val[u]%MOD;
    dis[++cnt] = w,id[cnt] = u;
    int v;
    for(int i = head[u];i != -1;i = edg[i].nxt)
    {
        v = edg[i].v;
        if(v == f || vis[v])
            continue;
        getdis(v,u,w);
    }
}
inline void solve(int u,int ssize)
{
    cc = 0,vis[u] = 1;
    visdis[val[u]%MOD] = 1;
    visid[val[u]%MOD] = u;
    cdis[++cc] = val[u]%MOD;
    int v;
    for(int i = head[u];i != -1;i = edg[i].nxt)
    {
        v = edg[i].v;
        if(vis[v]) continue;
        cnt = 0;
        getdis(v,v,val[u]%MOD);
        for(int i = 1;i <= cnt;++i)
        {
            int tmp = 1ll*inv[dis[i]]*k%MOD*val[u]%MOD;
            if(visdis[tmp])
            {
                int uu = visid[tmp],vv = id[i];
                if(uu > vv)
                    swap(vv,uu);
                if(uu < ans1 || (uu == ans1 && vv < ans2))
                    ans1 = uu,ans2 = vv;
            }
        }
        for(int i = 1;i <= cnt;++i)
        {
            if(!visdis[dis[i]])
                visdis[dis[i]] = 1,visid[dis[i]] = id[i],cdis[++cc] = dis[i];
            else if(id[i] < visid[dis[i]])
                visid[dis[i]] = id[i];
        }
    }
    for(int i = 1;i <= cc;++i)
        visdis[cdis[i]] = 0;
    for(int i = head[u];i != -1;i = edg[i].nxt)
    {
        v = edg[i].v;
        if(vis[v]) continue;
        Size = sz[v] < sz[u]?sz[v]:ssize-sz[u];
        mx = INF;
        getroot(v,v);
        solve(root,Size);
    }
}
inline void init()
{
    tot = 0,mx = INF,Size = n,ans1 = ans2 = INF;
    memset(vis,false,sizeof(bool) * (n+1));
    memset(head,-1,sizeof(int)*(n+1));
}
int main()
{
    getInv();
    while(~scanf("%d%d",&n,&k))
    {
        init();
        for(int i = 1;i <= n;++i)
            scanf("%d",&val[i]);
        int u,v;
        for(int i = 1;i < n;++i)
        {
            scanf("%d%d",&u,&v);
            addedg(u,v),addedg(v,u);
        }
        getroot(1,1);
        solve(root,Size);
        if(ans1 == INF)
            printf("No solution\n");
        else
            printf("%d %d\n",ans1,ans2);
    }
    return 0;
}