hdu4812 D Tree

树上的点分治。

首先为了避免超时预处理逆元，对于x(<mod)，我们希望找到f^-1(x) = y 满足y * x % mod = k。

由存在a, b : x *a + y * b = gcd(x, y), mod 为素数，故有 x * a + mod * b = 1。

由EXTENDED-EUCLID找出a (< mod)。

对于树上的分治，首先找出当前树的重心，处理剔除重心后的各个子树，合法链属于其中一个子树或者经过树根。

acm.hdu.edu.cn/showproblem.php?pid=4812

#include <cstdio>
#include <cstring>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
typedef __int64 LL;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 10;
const int mod = 1e6 + 3;

struct Edge{
    int to, next;
}edge[maxn * 2];
int head[maxn], f[mod], belong[mod], g[mod];
bool vis[maxn];
int n, N, root;
LL K, val[maxn], path[maxn];
int cnt1, mini;
LL d, x, y;
int ans[2], sum[maxn], id[maxn];

void addEdge(int u, int v){
    edge[N].next = head[u];
    edge[N].to = v;
    head[u] = N++;
}

void eEuclid(LL a, LL b){
    if(!b){
        d = a; x = 1; y = 0;
        return;
    }
    eEuclid(b, a % b);
    LL x1 = x;
    x = y;
    y = x1 - a / b * y;
}

void init(){
    //calculate inverse element
    for(int i = 0; i < mod; i++){
        eEuclid(mod, i);
        while(y < 0) y += mod;
        g[i] = y % mod;
    }
}

void getRoot(int u){
    sum[u] = 1;
    int maxii = 0;
    vis[u] = 1;
    for(int i = head[u]; i + 1; i = edge[i].next){
        int v = edge[i].to;
        if(vis[v]) continue;
        getRoot(v);
        sum[u] += sum[v];
        maxii = max(sum[v], maxii);
    }
    vis[u] = 0;
    maxii = max(maxii, sum[0] - sum[u]);
    if(maxii < mini) mini = maxii, root = u;
}

void update(int a, int b){
    if(a > b) swap(a, b);
    if(ans[0] == -1 || ans[0] > a){
        ans[0] = a, ans[1] = b;
        return;
    }
    if(ans[0] == a && ans[1] > b) ans[1] = b;
}

void dfs(int u, LL num){
    path[cnt1] = num * val[u] % mod;
    id[cnt1++] = u;
    LL tem = path[cnt1 - 1];
    vis[u] = 1;
    for(int i = head[u]; i + 1; i = edge[i].next){
        int v = edge[i].to;
        if(vis[v]) continue;
        dfs(v, tem);
    }
    vis[u] = 0;
}

void solve(int u, int cnt){
    if(cnt == 1) return;
    sum[0] = cnt;
    mini = inf;
    getRoot(u);
    vis[root] = 1;
    for(int i = head[root]; i + 1; i = edge[i].next){
        int v = edge[i].to;
        if(vis[v]) continue;
        cnt1 = 0;
        dfs(v, 1);
        for(int j = 0; j < cnt1; j++){
            if(path[j] * val[root] % mod == K) update(id[j], root);
            LL tem = K * g[path[j] * val[root] % mod] % mod;
            if(belong[tem] != N) continue;
            update(id[j], f[tem]);
        }
        for(int j = 0; j < cnt1; j++){
            int tem = path[j];
            if(belong[tem] != N || f[tem] > id[j]) f[tem] = id[j], belong[tem] = N;
        }
    }
    N++;
    for(int i = head[root]; i + 1; i = edge[i].next){
        int v = edge[i].to;
        if(vis[v]) continue;
        solve(v, sum[v]);
    }
}

int main(){
    init();
    while(~scanf("%d%I64d", &n, &K)){
        for(int i = 1; i <= n; i++) scanf("%I64d", &val[i]);
        memset(head, -1, sizeof head);
        memset(belong, -1, sizeof belong);
        memset(ans, -1, sizeof ans);
        memset(vis, 0, sizeof vis);
        N = 0;
        for(int i = 1, p, q; i < n; i++){
            scanf("%d%d", &p, &q);
            addEdge(p, q);
            addEdge(q, p);
        }
        N = 0;
        solve(1, n);
        if(ans[0] == -1) puts("No solution");
        else printf("%d %d
", ans[0], ans[1]);
    }
    return 0;
}