思路
圖論，最簡單的解法：
LCA加路徑長度判斷不等式
程式碼
#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
int f[N][25], d[N], dis[N], T, n, m, tot, t, ver[2 * N], next1[2 * N], head[N];
queue
void add(int x, int y) {
ver[++tot] = y;
next1[tot] = head[x];
head[x] = tot;
}
int lca(int x, int y) {
if (d[x] > d[y]) swap(x, y);
for (int i = t; i >= 0; i--) if (d[f[y][i]] >= d[x]) y = f[y][i];
if (x == y) return x;
for (int i = t; i >= 0; i--) if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
return f[x][0];
}
int dist(int a, int b) {
return dis[a] + dis[b] - 2 * dis[lca(a, b)];
}
int main() {
scanf("%d%d", &n, &m);
t = (int)(log(n) / log(2)) + 1;
for (int i = 1; i <= n; i++) head[i] = d[i] = 0;
for (int i = 1; i < n; i++) {
int x, y;
scanf("%d%d", &x, &y);
add(x, y);
add(y, x);
}
q.push(1);
d[1] = 1;
while (q.size()) {
int x = q.front();
q.pop();
for (int i = head[x]; i; i = next1[i]) {
int y = ver[i];
if (d[y]) continue;
d[y] = d[x] + 1;
dis[y] = dis[x] + 1;
f[y][0] = x;
for (int j = 1; j <= t; j++)
f[y][j] = f[f[y][j - 1]][j - 1];
q.push(y);
}
}
for (int i = 1; i <= m; i++) {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (dist(x1, y1) + dist(x2, y2) >= dist(x1, x2) + dist(y1, y2)) printf("Y\n");
else printf("N\n");
}
return 0;
}