HDU 4812 D Tree 树分治+逆元处理

D Tree

Problem Description

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 10⁶ + 3) equals to K?
Can you help them in solving this problem?

Input

There are several test cases, please process till EOF.
Each test case starts with a line containing two integers N(1 <= N <= 10⁵) and K(0 <=K < 10⁶ + 3). The following line contains n numbers v_i(1 <= v_i < 10⁶ + 3), where vi indicates the integer on vertex i. Then follows N - 1 lines. Each line contains two integers x and y, representing an undirected edge between vertex x and vertex y.

Output

For each test case, print a single line containing two integers a and b (where a < b), representing the two endpoints of the chain. If multiply solutions exist, please print the lexicographically smallest one. In case no solution exists, print “No solution”(without quotes) instead.
For more information, please refer to the Sample Output below.

Sample Input

5 60 2 5 2 3 3 1 2 1 3 2 4 2 5 5 2 2 5 2 3 3 1 2 1 3 2 4 2 5

Sample Output

3 4 No solution

Hint

1. “please print the lexicographically smallest one.”是指: 先按照第一个数字的大小进行比较，若第一个数字大小相同，则按照第二个数字大小进行比较，依次类推。 2. 若出现栈溢出，推荐使用C++语言提交，并通过以下方式扩栈： #pragma comment(linker,"/STACK:102400000,102400000")

题意：

　　给你一棵树n个点,一个K

　　让你找到一条 a->b 的字典数最小的路径满足这条路径上点权乘积取mod下等于K

题解：

　　预处理小于mod的所有逆元

　　树分治即可

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair

typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 1e5+10, M = 1e6+11, inf = 2e9, mod = 1000003;

int head[N],vis[N],f[N],siz[N],id[N],n,t = 1,ansl,ansr,allnode,root;

struct edge{int to,next;}e[N * 4];
LL mp[M],inv[M],v[M],K,deep[M];
void add(int u,int v) {e[t].next=head[u];e[t].to=v;head[u]=t++;}

void getroot(int u,int fa) {
        f[u] = 0;
        siz[u] =  1;
        for(int i = head[u]; i; i = e[i].next) {
            int to = e[i].to;
            if(vis[to] || to == fa) continue;
            getroot(to,u);
            siz[u] += siz[to];
            f[u] = max(f[u],siz[to]);
        }
        f[u] = max(f[u], allnode - siz[u]);
        if(f[u] < f[root]) root = u;
}
void getdeep(int u,int fa,LL now) {
    deep[++deep[0]] = now*v[u]%mod;
    id[deep[0]] = u;
    for(int i = head[u]; i; i = e[i].next) {
        int to = e[i].to;
        if(vis[to] || to == fa) continue;
        getdeep(to,u,now*v[u]%mod);
    }
}
void update(int u,int x,int y) {
        int tmp = mp[inv[x*v[u]%mod]*K%mod];
        if(!tmp) return ;
        if(y > tmp) swap(y,tmp);
        if(y < ansl || (y == ansl && tmp < ansr)) ansl = y, ansr = tmp;
}

void work(int u){
        vis[u] = 1;
        mp[1] = u;
        for(int i = head[u]; i; i = e[i].next) {
            int to = e[i].to;
            if(vis[to]) continue;
            deep[0] = 0;
            getdeep(to,u,1);
            for(int j = 1; j <= deep[0]; ++j) update(u,deep[j],id[j]);
            for(int j = 1; j <= deep[0]; ++j) if(!mp[deep[j]] || mp[deep[j]] > id[j])mp[deep[j]] = id[j];
        }
        mp[1] = 0;
         for(int i = head[u]; i; i = e[i].next) {
            int to = e[i].to;
            if(vis[to]) continue;
            deep[0] = 0;
            getdeep(to,u,1);
            for(int j = 1; j <= deep[0]; ++j)  mp[deep[j]] = 0;
        }
         for(int i = head[u]; i; i = e[i].next) {
            int to = e[i].to;
            if(vis[to]) continue;
            root = 0;
            allnode = siz[to];
            getroot(e[i].to,root);
            work(root);
         }
}
int main() {
    inv[1]=1;
    for(int i=2;i<mod;i++){int a=mod/i,b=mod%i;inv[i]=(inv[b]*(-a)%mod+mod)%mod;}
    while(~scanf("%d%I64d",&n,&K)) {
        t = 1;memset(head,0,sizeof(head));
        memset(vis,0,sizeof(vis));
        ansl = ansr = inf;
        for(int i = 1; i <= n; ++i) scanf("%I64d",&v[i]);
        for(int i = 1; i < n; ++i) {
            int u,v;
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        f[0]=inf;
        allnode=n;root=0;
        getroot(1,0);
        work(root);
        if(ansl == inf) puts("No solution");else
        printf("%d %d
",ansl,ansr);
    }
    return 0;
}