题意

三倍经验哇咔咔

Sol

首先可以证明，两点之间边权最大值最小的路径一定是在最小生成树上

考虑到这题是边权的最大值，直接把重构树建出来

然后查LCA处的权值即可

输入文件过大，需要用RMQ算法求LCA

// luogu-judger-enable-o2#include
    
     const int MAXN = 1e6 + 10;using namespace std;inline int read() {    char c = getchar(); int x = 0, f = 1;    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();    return x * f;}int N, Q, S, tot, dfn[MAXN], rev[MAXN], dep[MAXN], id[MAXN][21], lg2[MAXN], rd[MAXN];vector
     
       v[MAXN];void dfs(int x, int fa) {    dfn[x] = ++tot; dep[x] = dep[fa] + 1; id[tot][0] = x;     for(int i = 0, to; i < v[x].size(); i++) {        if((to = v[x][i]) == fa) continue;        dfs(to, x);        id[++tot][0] = x;    }}void RMQ() {    for(int i = 2; i <= tot; i++) lg2[i] = lg2[i >> 1] + 1;    for(int j = 1; j <= 20; j++) {        for(int i = 1; (i + (1 << j) - 1) <= tot; i++) {            int r = i + (1 << (j - 1));            id[i][j] = dep[id[i][j - 1]] < dep[id[r][j - 1]] ? id[i][j - 1] : id[r][j - 1];        }    }}int Query(int l, int r) {    if(l > r) swap(l, r);    int k = lg2[r - l + 1];    return dep[id[l][k]] < dep[id[r - (1 << k) + 1][k]] ? id[l][k] : id[r - (1 << k) + 1][k];}int main() {    freopen("a.in", "r", stdin);    N = read(); Q = read(); S = read();    for(int i = 1; i <= N - 1; i++) {        int x = read(), y = read();        v[x].push_back(y); v[y].push_back(x);    }    dfs(S, 0);    RMQ();    while(Q--) {        int x = read(), y = read();        printf("%d\n", Query(dfn[x], dfn[y]));    }    return 0;}