【HDU 4812】D Tree【树的分治】

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题意：给出一棵树，让你寻找一条路径，使得路径上的点相乘mod10^6+3等于k，输出路径的两个端点，按照字典序最小输出。

思路：这类问题很容易想到树的分治，每次找出树的重心，以重心为根，将树分成若干棵子树，然后对于每棵子树再一样的操作，现在就需要求一重心为根，寻找路径，依次遍历每一个子树，然后记录子树中点到根的权值的乘积X，然后通过在哈希表中寻找K×逆元(x)，看是否存在，存在则更新答案。

#pragma comment(linker,"/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define mod 1000003#define N 100005typedef long long ll;struct E{    int v, d, ne;    E() {}    E(int _v, int _ne):v(_v), ne(_ne){}}e[N*2];bool vis[N];int size, head[N], ans[2], root, flag[mod], F[mod], sum[N], mi, cr, id[N];ll val[N], ni[mod], path[N];void init() {    size = 0;    memset(vis, false, sizeof(vis));    memset(ans, -1, sizeof(ans));    memset(head, -1, sizeof(head));    memset(flag, 0, sizeof(flag));}void add(int u, int v) {    e[size] = E(v, head[u]);    head[u] = size++;}void dfs(int u, ll k) {    int i, v;    sum[u] = 1;    vis[u] = true, id[cr] = u;    path[cr++] = k*val[u]%mod;    ll tm = path[cr-1];    for (i = head[u];~i;i = e[i].ne) {        v = e[i].v;        if (vis[v]) continue;        dfs(v, tm);        sum[u] += sum[v];    }    vis[u] = false;}ll k;int n, ca;void getans(int a, int b) {    if (a > b) swap(a,b);    if (ans[0] == -1 || ans[0] > a) ans[0] = a, ans[1] = b;    else if (ans[0] == a && ans[1] > b) ans[1] = b;}void getroot(int u) {    int i, v, mx = 0;    sum[u] = 1;    vis[u] = true;    for (i = head[u];~i;i = e[i].ne) {        v = e[i].v;        if (vis[v]) continue;        getroot(v);        sum[u] += sum[v];        mx = max(mx, sum[v]);    }    mx = max(mx, sum[0]-sum[u]);    if (mx < mi) mi = mx, root = u;    vis[u] = false;}void cal(int u, int cnt) {    if (cnt == 1) return;    int i, v, j;    mi = n;    sum[0] = cnt;    getroot(u);    vis[root] = true;    for (i = head[root];~i;i = e[i].ne) {        v = e[i].v;        if (vis[v]) continue;        cr = 0;        dfs(v, 1);        for (j = 0;j < cr;j++) {            if (path[j]*val[root]%mod == k) getans(root, id[j]);            ll tm = k*ni[path[j]*val[root]%mod]%mod;            if (flag[tm] != ca) continue;            getans(F[tm], id[j]);        }        for (j = 0;j < cr;j++) {            int tm = path[j];            if (flag[tm] != ca || F[tm] > id[j]) F[tm] = id[j], flag[tm] = ca;        }    }    ca++;    for (i = head[root];~i;i = e[i].ne) {        if (vis[e[i].v]) continue;        cal(e[i].v, sum[e[i].v]);    }}ll egcd(ll a,ll b, ll &x, ll &y) {//得到的是a*x+b*y=gcd(a,b)ll temp,tempx;if (b == 0) {x = 1;y = 0;return a;}temp = egcd(b,a % b, x, y);tempx = x;x = y;y = tempx - a / b * y;return temp;}int main() {    int u, v, i, j;    ll y;    for (i = 0;i < mod;i++) {        egcd(i*1ll, mod*1ll, ni[i], y);        ni[i] %= mod, ni[i] = (ni[i]+mod)%mod;    }    while (~scanf("%d%I64d", &n, &k)) {        init();        ca = 1;        for (i = 1;i <= n;i++) scanf("%I64d", &val[i]);        for (i = 1;i < n;i++) {            scanf("%d%d", &u, &v);            add(u, v), add(v, u);        }        cal(1, n);        if (ans[0] == -1) puts("No solution");        else printf("%d %d\n", ans[0], ans[1]);    }}

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