BZOJ1832: [AHOI2008]聚会(LCA)

题目：

1832: [ahoi2008]聚会

解析：

偶尔做做水题挺爽的
两两之间先求出lca，发现至少有两个lca是相同的，这个重复lca也是深度最浅的那个，那我们就选择那个不重复的lca，因为若选这个重复的lca的话，这个重复的lca到另一个lca的路径会走两遍，反之只会走一遍
三点间的距离就是
\(dis[x] + dis[y] + dis[z] - dis[lca(x, y)]- dis[lca(x, z)]- dis[lca(y, z)]\)

代码：

#include <bits/stdc++.h>
using namespace std;
const int n = 1e6 + 10;
int n, m, num;
int head[n], dep[n], f[n], size[n], top[n], son[n];
struct node {
    int v, nx;
} e[n];

template<class t>inline void read(t &x) {
    x = 0; int f = 0; char ch = getchar();
    while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    x = f ? -x : x;
    return ;
}

inline void add(int u, int v) {
    e[++num].nx = head[u], e[num].v = v, head[u] = num;
}

void dfs1(int u, int fa) {
    size[u] = 1;
    for (int i = head[u]; ~i; i = e[i].nx) {
        int v = e[i].v;
        if (v != fa) {
            dep[v] = dep[u] + 1;
            f[v] = u;
            dfs1(v, u);
            size[u] += size[v];
            if (size[v] > size[son[u]]) son[u] = v;
        }
    }
}

void dfs2(int u, int t) {
    top[u] = t;
    if (son[u]) dfs2(son[u], t);
    for (int i = head[u]; ~i; i = e[i].nx) {
        int v = e[i].v;
        if (v != f[u] && v != son[u]) dfs2(v, v);
    }
}

int lca(int x, int y) {
    int fx = top[x], fy = top[y];
    while (fx != fy) {
        if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
        x = f[fx], fx = top[x];
    }
    return dep[x] < dep[y] ? x : y;
}

int main() {
    memset(head, -1, sizeof(head));
    read(n), read(m);
    for (int i = 1, x, y; i < n; ++i) read(x), read(y), add(x, y), add(y, x);
    f[1] = 0, dep[1] = 1;
    dfs1(1, 0);
    dfs2(1, 1);
    for (int i = 1, x, y, z; i <= m; ++i) {
        read(x), read(y), read(z);
        int a = lca(x, y), b = lca(x, z), c = lca(y, z);
        int tmp = (dep[a] == dep[b]) ? c : ((dep[a] == dep[c]) ? b : a);
        printf("%d %d\n", tmp, dep[x] + dep[y] + dep[z] - dep[a] - dep[b] - dep[c]);
    }
}