可以看出，三个点两两之间的lca会有一对相同，而另一个lca就是聚集点。
然后搞搞就可以求出距离了。
——代码
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #define MAXN 1000001 5 6 using namespace std; 7 8 int n, m, cnt, ans; 9 int head[MAXN], to[MAXN], next[MAXN], deep[MAXN], f[MAXN][21]; 10 11 inline void add(int x, int y) 12 { 13 to[cnt] = y; 14 next[cnt] = head[x]; 15 head[x] = cnt++; 16 } 17 18 inline void dfs(int u) 19 { 20 int i, v; 21 deep[u] = deep[f[u][0]] + 1; 22 for(i = 0; f[u][i]; i++) f[u][i + 1] = f[f[u][i]][i]; 23 for(i = head[u]; i != -1; i = next[i]) 24 { 25 v = to[i]; 26 if(!deep[v]) f[v][0] = u, dfs(v); 27 } 28 } 29 30 inline int lca(int x, int y) 31 { 32 int i; 33 if(deep[x] < deep[y]) swap(x, y); 34 for(i = 20; i >= 0; i--) 35 if(deep[f[x][i]] >= deep[y]) 36 x = f[x][i]; 37 if(x == y) return x; 38 for(i = 20; i >= 0; i--) 39 if(f[x][i] != f[y][i]) 40 x = f[x][i], y = f[y][i]; 41 return f[x][0]; 42 } 43 44 int main() 45 { 46 int i, x, y, z, a; 47 scanf("%d %d", &n, &m); 48 memset(head, -1, sizeof(head)); 49 for(i = 1; i < n; i++) 50 { 51 scanf("%d %d", &x, &y); 52 add(x, y); 53 add(y, x); 54 } 55 dfs(1); 56 for(i = 1; i <= m; i++) 57 { 58 scanf("%d %d %d", &x, &y, &z); 59 a = lca(x, y) ^ lca(x, z) ^ lca(y, z); 60 ans = deep[x] + deep[a] - 2 * deep[lca(x, a)]; 61 ans += deep[y] + deep[a] - 2 * deep[lca(y, a)]; 62 ans += deep[z] + deep[a] - 2 * deep[lca(z, a)]; 63 printf("%d %d ", a, ans); 64 } 65 return 0; 66 }