题意：求树上A，B两点路径上第K小的数
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<cstdlib>
#include<queue>
#include<set>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define eps 1e-4
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
typedef long long LL;
typedef long long ll;
const int maxn = 2e5 + 5;
const int mod = 998244353;
int n,q,m,TOT;
int a[maxn],t[maxn],T[maxn],lson[maxn*30],rson[maxn*30],c[maxn*30];
void Init_hash() {
for (int i = 1; i <= n; i++) t[i] = a[i];
sort(t + 1, t + 1 + n);
m = unique(t + 1, t + 1 + n) - t - 1;
}
int build(int l,int r) {
int root = TOT++;
c[root] = 0;
if (l != r) {
int mid = (l + r) >> 1;
lson[root] = build(l, mid);
rson[root] = build(mid + 1, r);
}
return root;
}
int Hash(int x) {
return lower_bound(t + 1, t + 1 + m, x) - t;
}
int update(int root,int pos,int val)
{
int newroot = TOT ++,tmp = newroot;
c[newroot] = c[root] + val;
int l = 1,r = m;
while(l <r)
{
int mid = (l+r)>>1;
if(pos <= mid)
{
lson[newroot] = TOT++;
rson[newroot] = rson[root];
newroot = lson[newroot];
root = lson[root];
r = mid;
}
else
{
rson[newroot] = TOT ++;
lson[newroot] = lson[root];
newroot = rson[newroot];
root = rson[root];
l = mid + 1;
}
c[newroot] = c[root] + val;
}
return tmp;
}
int query(int left_root,int right_root,int LCA,int k) {
int lca_root = T[LCA];
int pos = Hash(a[LCA]);
int l = 1, r = m;
while (l < r) {
int mid = (l + r) >> 1;
int tmp = c[lson[left_root]] + c[lson[right_root]] - 2 * c[lson[lca_root]] + (pos >= l && pos <= mid);
if (tmp >= k) {
left_root = lson[left_root];
right_root = lson[right_root];
lca_root = lson[lca_root];
r = mid;
} else {
k -= tmp;
left_root = rson[left_root];
right_root = rson[right_root];
lca_root = rson[lca_root];
l = mid + 1;
}
}
return l;
}
//*********************LAC**************************
int rmq[2 * maxn]; // 欧拉序列对应的深度序列
struct ST {
int mm[2 * maxn];
int dp[2 * maxn][20];
void init(int n) {
mm[0] = -1;
for (int i = 1; i <= n; i++) {
mm[i] = ((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
dp[i][0] = i;
}
for (int j = 1; j <= mm[n]; j++)
for (int i = 1; i + (1 << j) - 1 <= n; i++)
dp[i][j] =
rmq[dp[i][j - 1]] < rmq[dp[i + (1 << (j - 1))][j - 1]] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][
j - 1];
}
int query(int a, int b) {
if (a > b) swap(a, b);
int k = mm[b - a + 1];
return rmq[dp[a][k]] <= rmq[dp[b - (1 << k) + 1][k]] ? dp[a][k] : dp[b - (1 << k) + 1][k];
}
}st;
struct Edge {
int to, next;
}edge[maxn * 2];
int tot,head[maxn * 2];
int F[maxn * 2]; //欧拉序列 即dfs遍历的顺序
int P[maxn]; //表示点i在F中第一次出现的位置
int cnt;
void init() {
tot = 0;
memset(head, -1, sizeof head);
}
void addedge(int u,int v) {
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void dfs(int u,int pre,int dep) {
F[++cnt] = u;
rmq[cnt] = dep;
P[u] = cnt;
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if (v == pre) continue;
dfs(v, u, dep + 1);
F[++cnt] = u;
rmq[cnt] = dep;
}
}
void LCA_init(int root,int node_num) {
cnt = 0;
dfs(root, root, 0);
st.init(2 * node_num - 1);
}
int query_lca(int u,int v) {
return F[st.query(P[u], P[v])];
}
void dfs_build(int u,int pre) {
int pos = Hash(a[u]);
T[u] =update(T[pre],pos,1);
for(int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if(v == pre) continue;
dfs_build(v,u);
}
}
int main() {
while(scanf("%d %d",&n,&q) == 2) {
for(int i = 1; i <= n; i++) scanf("%d",&a[i]);
Init_hash();
init();
TOT = 0;
int u,v;
for(int i = 1;i < n; i ++) {
scanf("%d %d",&u,&v);
addedge(u,v);
addedge(v,u);
}
LCA_init(1,n);
T[n + 1] = build(1,m);
dfs_build(1,n + 1);
int k;
while(q--) {
scanf("%d %d %d",&u,&v,&k);
int ans = t[query(T[u],T[v],query_lca(u,v),k)];
printf("%d\n",ans);
}
return 0;
}
return 0;
}
来源：https://www.cnblogs.com/smallhester/p/11423216.html