【倍增求lca】小机房的树

2370 小机房的树

//讲真我觉得这题面有点恶心QAQ

大体思路：树上的最短路径一定经过两点lca， 预处理一个dis数组， 存每个点到根节点的距离， 两点（x， y） 之间的距离即为dis[x] + dis[y] - 2*dis[lca] ；

然后倍增求lca即可，dis数组在预处理每个点深度的时候一并处理

日常，“妙啊”

代码君qwq

#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
const int maxn = 500050, maxm = 500050;
int n, m, s, num = 0, log2n;
int head[maxm], jump[maxn][23], dep[maxn], dis[maxn];
struct edge {
    int nxt, to, dis;
}e[maxm<<1];
void add(int from, int to, int dis) {
    e[++num].nxt = head[from];
    e[num].to = to;
    e[num].dis = dis;
    head[from] = num;
}
void dfs(int v) {
    for(int i = head[v]; i; i = e[i].nxt) {
        int u = e[i].to;
        if(jump[v][0] != u) {
            jump[u][0] = v;
            dep[u] = dep[v] + 1;
            dis[u] = dis[v]+e[i].dis;
            dfs(u);
        }
    }
}
void init() {
/**/for(int i = 1; i <= log2n; i++) 
        for(int j = 1; j <= n; j++) 
            jump[j][i] = jump[jump[j][i-1]][i-1];
}
int LCA(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y);
    int t = dep[x] - dep[y];
    for(int i = 0; i <= log2n; i++) {
        if(t&(1<<i)) x = jump[x][i];
    }
    if(x == y) return x;
    for(int i = log2n; i >= 0; i--) {
        if(jump[x][i] != jump[y][i]) {
            x = jump[x][i];
            y = jump[y][i];
        }
    }
    return jump[x][0];
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i < n; i++) {
        int x, y, z;
        scanf("%d%d%d", &x, &y, &z);
        add(x, y, z), add(y, x, z);
    }    
    s = 0;
    log2n = log(n)/log(2)+1;
    dep[s] = 1;
    jump[s][0] = 0;
    dis[s] = 0;
    dfs(s);
    init();
    scanf("%d", &m);
    for(int i = 1; i <= m; i++) {
        int x, y, ans = 0, lca;
        scanf("%d%d", &x, &y);
        lca = LCA(x, y);
        ans = dis[x] + dis[y] - 2*dis[lca];
        printf("%d
", ans);
    }
    return 0;
}