c/c++语言开发共享LOJ#137. 最小瓶颈路 加强版(Kruskal重构树 rmq求LCA)

题意

三倍经验哇咔咔

#6021. 「from commonants」寻找 lcr

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<bits/stdc++.h> 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<int> 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("%dn", query(dfn[x], dfn[y]));     }     return 0; }