樹上跳棋

題目鏈接

`戳我

\(Solution\)

對於一個點如果能夠被跳到當且僅當這個點的深度\(mod\)一次跳的長度等於起始節點\(mod\)一次跳的長度
假設能夠被\(p1,p2\)兩個點都能到達的點爲\(z\)需要滿足以下條件

\[dep[z]<=dep[lca] \]

\[dep[z]\equiv dep[p1]\ (mod \ d1) \]

\[dep[z]\equiv dep[p2]\ (mod \ d2) \]

下面那兩個同餘的限制可以用\(excrt\)(擴展中國剩餘定理)算出最小的\(dep[z]\)
然後滿足的\(dep[z]\)就是\(dep[z]+k*lcm(d1,d2)\quad k \in N^*\)
通過這個可以算出滿足\(dep[z]<=dep[lca]\)最大的\(dep[z]\)即兩個棋子的最近的重合節點
對於\(-1\)的情況即需要判斷一下是否有解和是否存在\(dep[x]<=dep[lca]\)即可

\(Code\)

#include <bits/stdc++.h>
#define int long long
using namespace std;
typedef long long ll;
int read() {
    int x = 0, f = 1;
    char c = getchar();
    while(c < '0' || c > '9')
        f = (c == '-') ? -1 : 1, c = getchar();
    while(c >= '0' && c <= '9')
        x = x * 10 + c - 48, c = getchar();
    return f * x;
}
struct node {
    int to, next;
} a[400010];
int head[200010], cnt;
int f[200010][21], dep[200010];
void add(int x, int y) {
    a[++cnt].next = head[x];
    head[x] = cnt;
    a[cnt].to = y;
}
void dfs(int x, int fa) {
    f[x][0] = fa;
    dep[x] = dep[fa] + 1;
    for(int i = 1; i <= 20; i++)
        f[x][i] = f[f[x][i - 1]][i - 1];
    for(int i = head[x]; i; i = a[i].next) {
        int v = a[i].to;
        if(v == fa)
            continue;
        dfs(v, x);
    }
}
int lca(int x, int y) {
    if(dep[x] > dep[y])
        swap(x, y);
    for(int i = 20; i >= 0; i--)
        if(dep[f[y][i]] >= dep[x])
            y = f[y][i];
    if(x == y)
        return x;
    for(int i = 20; i >= 0; i--)
        if(f[x][i] != f[y][i])
            x = f[x][i], y = f[y][i];
    return f[x][0];
}
int jump(int x, int depth) {
    for(int i = 20; i >= 0; i--)
        if(dep[f[x][i]] >= depth)
            x = f[x][i];
    return x;
}
int exgcd(int a, int b, int& x, int& y) {
    if(!b) {
        x = 1, y = 0;
        return a;
    }
    int gcd = exgcd(b, a % b, x, y);
    int X = x, Y = y;
    x = Y, y = X - a / b * Y;
    return gcd;
}
int ksm(int a, int b, int p) {
    int ans = 0;
    while(b) {
        if(b & 1)
            ans = (ans + a) % p;
        a = (a + a) % p;
        b >>= 1;
    }
    return ans;
}
int m[10], A[10];
int find(int X, int Y, int a, int b) {
    A[1] = X, A[2] = Y;
    m[1] = a, m[2] = b;
    int lcm = m[1], ans = A[1] % m[1];
    int c = ((A[2] - ans) % m[2] + m[2]) % m[2], x, y;
    int gcd = exgcd(lcm, m[2], x, y);
    int k = ksm(x, c / gcd, m[2] / gcd);
    ans += lcm * k;
    lcm *= m[2] / gcd;
    ans = (ans % lcm + lcm) % lcm;
    return ans;
}
signed main() {
    int n = read(), x, y;
    for(int i = 1; i < n; i++)
        x = read(), y = read(), add(x, y), add(y, x);
    int q = read();
    dfs(1, 0);
    while(q--) {
        int x = read(), d1 = read(), y = read(), d2 = read();
        int Lca = lca(x, y), lcm = d1 * d2 / __gcd(d1, d2);
        int minx = find(dep[x] % d1, dep[y] % d2, d1, d2);
        if(minx == 0)
            minx += lcm;
        if(minx % d1 != dep[x] % d1 || minx % d2 != dep[y] % d2) {
            puts("-1");
            continue;
        }
        if(minx > dep[Lca]) {
            puts("-1");
            continue;
        }
        int depth = (dep[Lca] - minx) / lcm * lcm + minx;
        cout << jump(x, depth) << "\n";
    }
}