D Tree

题目描述

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?

题意概述

给定一棵有$N$个节点的树，其中第$i$个节点的权值为$v_i$。求一个字典序最小的点对使得这两点之间路径上的点权之积模$10^6+3$等于$K$。

数据范围：$1 \le N \le 10^5, \ 0 \le K \lt 10^6+3, \ 1 \le v_i \lt 10^6+3$。

算法分析

点分治。设当前枚举到的节点为$x$，其第$i$棵子树的根节点为$x_i$。用 map 储存下$x$前$i$棵子树中的节点到$x$儿子的可能的乘积。在计算第$(i+1)$棵子树时，设当前计算的节点到$x_{i+1}$的乘积为$p$，若 map 中存在$q$满足$p \times q \times x \equiv K \pmod {10^6+3}$，即$q \equiv K \times p^{-1} \times x^{-1} \pmod {10^6+3}$，则将这两个节点与当前答案比较，取较优解。

代码

#include <cstdio>
#include <map>
#include <algorithm>
#include <cstring>
#define MOD 1000003
using namespace std;
struct edge {
    int v, nxt;
} e[200001];
long long n, k, x, y, root, nume, tot, inv[MOD], h[100001], v[100001], size[100001], f[100001], val[100001];
bool vis[100001];
map<long long, int> id;
void add_edge(int u, int v) {
    e[++nume].v = v, e[nume].nxt = h[u], h[u] = nume;
    e[++nume].v = u, e[nume].nxt = h[v], h[v] = nume;
}
void get_root(int t, int fa) {
    size[t] = 1, f[t] = 0;
    for (int i = h[t]; i; i = e[i].nxt) {
        if (!vis[e[i].v] && e[i].v != fa) {
            get_root(e[i].v, t);
            size[t] += size[e[i].v];
            f[t] = max(f[t], size[e[i].v]);
        }
    }
    f[t] = max(f[t], tot - size[t]);
    if (f[t] < f[root]) root = t;
}
void update(int a, int b) {
    if (a > b) swap(a, b);
    if (a < x) x = a, y = b;
    else if (a == x && b < y) y = b;
}
void get_dist(int t, int fa, int flag) {
    if (!flag) {
        if (!id.count(val[t])) id[val[t]] = t;
        else id[val[t]] = min(id[val[t]], t);
    } else {
        if (val[t] * val[root] % MOD == k) {
            if (t <= x || root <= x) update(t, root);
        }
        long long inverse = k * inv[val[t]] % MOD * inv[val[root]] % MOD;
        if (id.count(inverse)) {
            if (id[inverse] <= x || t <= x) update(id[inverse], t);
        }
    }
    for (int i = h[t]; i; i = e[i].nxt) {
        if (!vis[e[i].v] && e[i].v != fa) {
            if (flag) val[e[i].v] = val[t] * v[e[i].v] % MOD;
            get_dist(e[i].v, t, flag);
        }
    }
}
void solve(int t) {
    vis[t] = true, val[t] = v[t], id.clear();
    for (int i = h[t]; i; i = e[i].nxt) {
        if (!vis[e[i].v]) {
            val[e[i].v] = v[e[i].v];
            get_dist(e[i].v, t, 1);
            get_dist(e[i].v, t, 0);
        }
    }
    for (int i = h[t]; i; i = e[i].nxt) {
        if (!vis[e[i].v]) {
            root = 0, tot = size[e[i].v];
            get_root(e[i].v, t);
            solve(root);
        }
    }
}
int main()
{
    inv[1] = 1;
    for (int i = 2; i < MOD; ++i) {
        inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
    }
    while (scanf("%lld%lld", &n, &k) != -1) {
        x = y = 1e9, nume = 0;
        memset(vis, 0, sizeof(vis));
        memset(h, 0, sizeof(h));
        for (int i = 1; i <= n; ++i) scanf("%lld", &v[i]);
        for (int i = 1; i < n; ++i) {
            int u, v;
            scanf("%d%d", &u, &v);
            add_edge(u, v);
        }
        tot = f[0] = n, root = 0;
        get_root(1, 0);
        solve(root);
        if (y <= n) printf("%lld %lld\n", x, y);
        else printf("No solution\n");
    }
    return 0;
}