Luogu Count on a tree(链上第k小)

Posted on |

Problem

给一颗n个节点的树，节点有权值，求以u，v为端点的条上权值第k小。

$n,m<1e5$询问在线。

Solution

显然要用主席树维护。

但是，很难将链转化为序列。

所以要一种巧妙的做法。

每个节点的主席树由它的父亲继承下来。

然后，查询链的时候，只需要减去u，v的LCA和LCA的父亲的贡献即可。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
#include <bits/stdc++.h>

using namespace std;

#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define fst first
#define snd second

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

inline int read(){
	int res = 0, fl = 1;
	char r = getchar();
	for (; !isdigit(r); r = getchar()) if(r == '-') fl = -1;
	for (; isdigit(r); r = getchar()) res = (res << 3) + (res << 1) + r - 48;
	return res * fl;
}
typedef long long LL;
typedef pair<int, int> pii;
const int Maxn = 1e5 + 10;
struct SGT{
	int ls,rs,sum;
}tre[Maxn << 6];
int val[Maxn], fa[Maxn][30], cnt, dep[Maxn];
int num, root[Maxn << 6], B[Maxn];
vector <int> g[Maxn];
int LCA(int u,int v){
	if(dep[u] < dep[v]) swap(u, v);
	for (int i = 18; i >= 0 ; --i) if(dep[fa[u][i]] >= dep[v]) u = fa[u][i];
	if(u == v) return u;
	for (int i = 18; i >= 0 ; --i) if(fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
	return fa[u][0];
}
void build(int &rt,int grt,int l,int r,int pos){
	tre[rt = ++cnt] = tre[grt];
	tre[rt].sum++;
	if(l == r) return;
	int mid = l + r >> 1;
	if(pos <= mid) build(tre[rt].ls, tre[grt].ls, l, mid, pos);
	else build(tre[rt].rs, tre[grt].rs, mid + 1, r, pos);
}
int Query(int a,int b,int c,int d,int l,int r,int k){
	if(l == r) return B[l];
	int mid = l + r >> 1, t = tre[tre[a].ls].sum + tre[tre[b].ls].sum - tre[tre[c].ls].sum - tre[tre[d].ls].sum;
	if(k <= t) return Query(tre[a].ls, tre[b].ls, tre[c].ls, tre[d].ls, l, mid, k);
	else return Query(tre[a].rs, tre[b].rs, tre[c].rs, tre[d].rs, mid + 1, r, k - t);
}
void dfs(int now, int pa){
	dep[now] = dep[pa] + 1;
	fa[now][0] = pa;
	for (int i = 1; i <= 18; ++i) fa[now][i] = fa[fa[now][i - 1]][i - 1];
	build(root[now],root[pa], 1, num, val[now]);
	for (int i = g[now].size() - 1; i >= 0; --i){
		int nxt = g[now][i];
		if(nxt == pa) continue;
		dfs(nxt, now);
	}
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in", "r", stdin);
	freopen("a.out", "w", stdout);
#endif
	int n = read(), m = read();
	for (int i = 1; i <= n; ++i) B[i] = val[i] = read();
	sort(B + 1, B + 1 + n);
	num = unique(B + 1, B + 1 + n) - B - 1;
	for (int i = 1; i <= n; ++i) val[i] = lower_bound(B + 1, B + num + 1, val[i]) - B;
	for (int i = 1; i < n; ++i) {
		int x = read(), y = read();
		g[x].push_back(y);
		g[y].push_back(x);
	}
	dfs(1,0);
	int ans = 0;
	while(m--){
		int u = read() ^ ans, v = read(), k = read();
		int lca = LCA(u,v),flca = fa[lca][0];
		ans = Query(root[u],root[v],root[lca],root[flca],1,num,k);
		printf("%d\n",ans);
	}
	return 0;
}