询问[l,r]的公共祖先,类似RMQ用dp[i][j]表示[i,i+(1<<j)-1]的公共祖先。
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = 300010;
const int DEG = 20;
struct Edge
{
int to,next;
}edge[MAXN*2];
int head[MAXN],tot;
void addedge(int u,int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void init()
{
tot = 0;
memset(head,-1,sizeof(head));
}
int fa[MAXN][DEG];//fa[i][j]表示结点i的第2^j个祖先
int deg[MAXN];//深度数组
void BFS(int root)
{
queue<int>que;
deg[root] = 0;
fa[root][0] = root;
que.push(root);
while(!que.empty())
{
int tmp = que.front();
que.pop();
for(int i = 1;i < DEG;i++)
fa[tmp][i] = fa[fa[tmp][i-1]][i-1];
for(int i = head[tmp]; i != -1;i = edge[i].next)
{
int v = edge[i].to;
if(v == fa[tmp][0])continue;
deg[v] = deg[tmp] + 1;
fa[v][0] = tmp;
que.push(v);
}
}
}
int LCA(int u,int v)
{
if(deg[u] > deg[v])swap(u,v);
int hu = deg[u], hv = deg[v];
int tu = u, tv = v;
for(int det = hv-hu, i = 0; det ;det>>=1, i++)
if(det&1)
tv = fa[tv][i];
if(tu == tv)return tu;
for(int i = DEG-1; i >= 0; i--)
{
if(fa[tu][i] == fa[tv][i])
continue;
tu = fa[tu][i];
tv = fa[tv][i];
}
return fa[tu][0];
}
int dp[MAXN][20];
int main()
{
int n;
while(scanf("%d",&n) == 1){
init();
int u,v;
for(int i = 1;i < n;i++){
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
}
BFS(1);
for(int i = 1;i <= n;i++)
dp[i][0] = i;
for(int j = 1;j < 20;j++){
for(int i = 1;i+(1<<j)-1 <= n;i++)
dp[i][j] = LCA(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int Q;
scanf("%d",&Q);
while(Q--){
scanf("%d%d",&u,&v);
int k=(int)log2(v-u+1);
printf("%d\n",LCA(dp[u][k],dp[v-(1<<k)+1][k]));
}
}
return 0;
}