题解【洛谷P2633】Count on a tree

题面

就是把静态区间第 K 小问题搬到了树上。。。可以先做一下 主席树模板。

我们知道主席树本质上是一棵权值线段树，具有可减性。

然后用树上差分那一套就行了，两点间查询时的权值线段树是 (rt_u+rt_v-rt_{ ext{LCA}(u, v)} - rt_{fa_{ ext{LCA(u, v)}}})。

代码有点长。。。

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

template <typename T>
inline T gi()
{
	T f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int INF = 0x3f3f3f3f, N = 100003, M = N << 1;

int n, m;
int a[N], b[N], lst, tott;
int tot, head[N], ver[M], nxt[M];
struct Node {int ls, rs, cnt;} tr[N * 40];
int rt[N * 40];
int sz[N], son[N], dep[N], fa[N], topp[N];

inline void add(int u, int v) {ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;}

struct LCA
{
	void dfs1(int u, int f)
	{
		fa[u] = f, sz[u] = 1, dep[u] = dep[f] + 1;
		for (int i = head[u]; i; i = nxt[i])
		{
			int v = ver[i];
			if (v == f) continue;
			dfs1(v, u);
			sz[u] += sz[v];
			if (sz[v] > sz[son[u]]) son[u] = v;
		}
	}
	void dfs2(int u, int f)
	{
		topp[u] = f;
		if (!son[u]) return;
		dfs2(son[u], f);
		for (int i = head[u]; i; i = nxt[i])
		{
			int v = ver[i];
			if (v == fa[u] || v == son[u]) continue;
			dfs2(v, v);
		}
	}
	int getLCA(int u, int v)
	{
		while (topp[u] != topp[v])
		{
			if (dep[topp[u]] < dep[topp[v]]) swap(u, v);
			u = fa[topp[u]];
		}
		if (dep[u] < dep[v]) return u;
		return v;
	}
} lca;

struct ChairmanTree
{
	int build(int l, int r)
	{
		int p = ++tott;
		if (l == r) return p;
		int mid = (l + r) >> 1;
		tr[p].ls = build(l, mid);
		tr[p].rs = build(mid + 1, r);
		return p;
	}
	inline void pushup(int p) 
	{
		tr[p].cnt = tr[tr[p].ls].cnt + tr[tr[p].rs].cnt;
	}
	int modify(int p, int val, int l, int r)
	{
		int q = ++tott; tr[q] = tr[p];
		if (l == r) {++tr[q].cnt; return q;}
		int mid = (l + r) >> 1;
		if (val <= mid) tr[q].ls = modify(tr[p].ls, val, l, mid);
		else tr[q].rs = modify(tr[p].rs, val, mid + 1, r);
		pushup(q);
		return q;
	}
	int query(int p, int q, int rr, int s, int k, int l, int r)
	{
		if (l == r) return l;
		int lcnt = tr[tr[p].ls].cnt + tr[tr[q].ls].cnt - tr[tr[rr].ls].cnt - tr[tr[s].ls].cnt;
		int mid = (l + r) >> 1;
		if (k <= lcnt) return query(tr[p].ls, tr[q].ls, tr[rr].ls, tr[s].ls, k, l, mid);
		else return query(tr[p].rs, tr[q].rs, tr[rr].rs, tr[s].rs, k - lcnt, mid + 1, r);
	}
} ct;

void dfs(int u, int f)
{
	rt[u] = ct.modify(rt[fa[u]], a[u], 1, n);
	for (int i = head[u]; i; i = nxt[i])
	{
		int v = ver[i];
		if (v == f) continue;
		dfs(v, u);
	}
}

int main()
{
	//File("");
    n = gi <int> (), m = gi <int> ();
    for (int i = 1; i <= n; i+=1) a[i] = b[i] = gi <int> ();
    sort(b + 1, b + 1 + n);
    int cntt = unique(b + 1, b + 1 + n) - b;
    for (int i = 1; i <= n; i+=1) 
	    a[i] = lower_bound(b + 1, b + 1 + cntt, a[i]) - b;
	for (int i = 1; i < n; i+=1)
	{
		int u = gi <int> (), v = gi <int> ();
		add(u, v), add(v, u);
	}
	lca.dfs1(1, 0); lca.dfs2(1, 1);
	rt[0] = ct.build(1, n);
	dfs(1, 0);
	while (m--)
	{
		int u = gi <int> () ^ lst, v = gi <int> (), k = gi <int> ();
		int l = lca.getLCA(u, v);
		printf("%d
", lst = b[ct.query(rt[u], rt[v], rt[l], rt[fa[l]], k, 1, n)]);
	}
	return 0;
}