【最近公共祖先】洛谷_3379 最近公共祖先（LCA）

题意

给出一棵节点数为$N$的树，有$M$次询问，每次询问两个节点的最近公共祖先。

思路

树上倍增。

代码

#include<queue>
#include<cmath>
#include<cstdio>
using namespace std;

struct node{
    int to, next;
}e[1000001];
int N, M, S, tot, t;
int d[500001], head[500001], f[500001][20];

void bfs(int S)
{
    int x, y;
    queue<int> q;
    d[S] = 1;
    q.push(S);
    while (q.size())
    {
        x = q.front();
        q.pop();
        for (int i = head[x]; i; i = e[i].next)
        {
            y = e[i].to;
            if (d[y]) continue;
            d[y] = d[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);
        }
    }
}

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 main()
{
    scanf("%d %d %d", &N, &M, &S);
    int x, y;
    t = log(N) / log(2);
    for (int i = 1; i < N; i++)
    {
        scanf("%d %d", &x, &y);
        e[++tot] = (node){y, head[x]};
        head[x] = tot;
        e[++tot] = (node){x, head[y]};
        head[y] = tot;
    }
    bfs(S);
    while (M--)
    {
        scanf("%d %d", &x, &y);
        printf("%d\n", LCA(x, y));
    }
}