题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1787
Description
Input
Output
LCA水过
三个点两两求LCA，若有其中两个LCA一样“集合点”就是第三个LCA
多加inline有用处，DFS改BFS对效率没有什么提高
从上到下依次是DFS（小改），BFS，DFS
DFS Version
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #define rep(i,l,r) for(int i=l; i<=r; i++) 6 #define clr(x,y) memset(x,y,sizeof(x)) 7 #define travel(x) for(Edge *p=last[x]; p; p=p->pre) 8 using namespace std; 9 const int INF = 0x3f3f3f3f; 10 const int maxn = 500010; 11 struct Edge{ 12 Edge *pre; int to; 13 }edge[maxn<<1]; 14 Edge *last[maxn],*pt; 15 int n,m,x,y,z,depth[maxn],fa[maxn][19]; 16 bool vis[maxn]; 17 inline int read(){ 18 int ans = 0, f = 1; 19 char c = getchar(); 20 while (!isdigit(c)){ 21 if (c == '-') f = -1; 22 c = getchar(); 23 } 24 while (isdigit(c)){ 25 ans = ans * 10 + c - '0'; 26 c = getchar(); 27 } 28 return ans * f; 29 } 30 inline void addedge(int x,int y){ 31 pt->pre = last[x]; pt->to = y; last[x] = pt++; 32 } 33 void dfs(int x){ 34 vis[x] = 1; 35 rep(i,1,18){ 36 if (depth[x] < (1 << i)) break; 37 fa[x][i] = fa[fa[x][i-1]][i-1]; 38 } 39 travel(x){ 40 if (vis[p->to]) continue; 41 depth[p->to] = depth[x] + 1; 42 fa[p->to][0] = x; 43 dfs(p->to); 44 } 45 } 46 inline int lca(int x,int y){ 47 if (depth[x] < depth[y]) swap(x,y); 48 int t = depth[x] - depth[y]; 49 rep(i,0,18) if (t & (1 << i)) x = fa[x][i]; 50 if (x == y) return x; 51 for(int i=18; i>=0; i--) if (fa[x][i] != fa[y][i]){ 52 x = fa[x][i]; y = fa[y][i]; 53 } 54 return fa[x][0]; 55 } 56 inline int calc(int x,int y){ 57 return depth[x] + depth[y] - (depth[lca(x,y)] << 1); 58 } 59 inline void solve(int x,int y,int z){ 60 int l1 = lca(x,y), l2 = lca(y,z), l3 = lca(x,z), t; 61 if (l1 == l2) t = l3; else if (l2 == l3) t = l1; else t = l2; 62 int ans = calc(x,t) + calc(y,t) + calc(z,t); 63 printf("%d %d\n",t,ans); 64 } 65 int main(){ 66 n = read(); m = read(); clr(last,0); pt = edge; 67 rep(i,1,n-1){ 68 x = read(); y = read(); addedge(x,y); addedge(y,x); 69 } 70 clr(vis,0); dfs(1); 71 rep(i,1,m){ 72 x = read(); y = read(); z = read(); solve(x,y,z); 73 } 74 return 0; 75 }