[HNOI2015] 开店

合集 - 题解(39)

1.P2048 [NOI2010] 超级钢琴 RMQ + Heap2022-03-312.【CDQ分治】 CF641E2022-03-313.生活在树上（hard version) 树上差分2022-04-194.【牛逼做法】P4186 Cow at Large2022-04-035.P3519 [POI2011]ROZ-Difference2023-07-026.【CF1146F】Leaf Partition2023-07-027.【CF1715E】Long Way Home2023-07-028.【CF678F】Lena and Queries2023-07-029.【CF1797F】Li Hua and Path2023-07-0210.P9170 填数游戏 贺题记录2023-07-0211.P7316 [COCI2018-2019#3] NLO2023-07-0212.P5471 [NOI2019] 弹跳2023-07-0213.【经典例题】P6822 [PA2012] Tax2023-07-0214.【差分 Trick】CF626F Group Projects2023-07-0315.CF1556G Gates to Another World2023-08-0616.【树论典题。】P6071 『MdOI R1』Treequery2023-08-0817.神秘哈希题。2023-08-0918.【树论，计数】Centroid Probabilities2023-08-1019.【Tricks，典】[ARC085F] NRE2023-08-1220.【拆贡献】CF1422F Boring Queries2023-08-2821.【OGF、Lucas】P4640 [BJWC2008] 王之财宝2023-08-2822.【Quick Hull】P3236 [HNOI2014] 画框2023-09-0823.【很难啊、拆分数、观察】P6944 [ICPC2018 WF] Gem Island2023-09-0324.「突刺贯穿第二分块」P4117 [Ynoi2018] 五彩斑斓的世界2023-09-0325. 【LCT 维护子树信息】CF916E Jamie and Tree2023-09-0226.【线段树合并、虚树】P5327 [ZJOI2019] 语言2023-09-1527.【树套树，LCT，出栈序】P4027 [NOI2007] 货币兑换2023-09-0928.【边双，dp】P8867 [NOIP2022] 建造军营2023-10-1329.【历史和】P8868 [NOIP2022] 比赛2023-10-1430.CF1083D2023-11-0631.CF1486F2023-11-06

32.[HNOI2015] 开店2023-11-06

33.【LCT、树状数组】CF1137F Matches Are Not a Child's Play2023-10-1934.【链交理论】CF1336F Journey2024-01-2635.【树论典题】P5642 人造情感（emotion）2023-11-1036.【友谊就是魔法！！】CF453E2024-01-2637.CF1109E、CF1109F2024-01-2638.P10083 [GDKOI2024 提高组] 不休陀螺2024-01-2639.洛谷博客杂题 #12024-02-27

收起

妈的，杀软动态点分治。

你考虑建出点分树，然后把所有子树塞进该点。

根据经典结论 $\sum dep_x = \sum sz_x = n\log n$ 然后我们考虑每次按照 $v$ 来排序，做前缀和，然后我们发现每次我们只需要查询一段区间和，使用二分查找即可。

注意容斥，具体来说，就是考虑在 $x$ 的时候容斥掉 $sub_x \to fa_x$ 这一类的贡献，可以在建树的时候计算 $dis_{t \to fa_x}$ 即可，然后我们用子树大小减去不合法的再乘上 $dis_{x \to fa_{pos}}$ 即可。

#include <bits/stdc++.h>
#define rep(i, l, r) for (int i = l; i <= r; i ++)
#define per(i, r, l) for (int i = r; i >= l; i --)

using namespace std;
typedef long long ll;

const int _ = 2e5 + 5, mod = 998244353, inf = 0x3f3f3f3f;

int power (int x, int y) {
	int ret = 1;
	for ( ; y; y >>= 1, x = 1ll * x * x % mod)
		if (y & 1) ret = 1ll * ret * x % mod;
	return ret;
}
void add (int & x, int y) { x = x + y >= mod ? x + y - mod : x + y; }
int mul (int x, int y) { return 1ll * x * y % mod; }

#define rep(i, l, r) for (int i = l; i <= r; i ++)
#define per(i, r, l) for (int i = r; i >= l; i --)

using namespace std;
typedef long long ll;

const int _ = 2e5 + 5, mod = 998244353, inf = 0x3f3f3f3f;

int power (int x, int y) {
	int ret = 1;
	for ( ; y; y >>= 1, x = 1ll * x * x % mod)
		if (y & 1) ret = 1ll * ret * x % mod;
	return ret;
}
void add (int & x, int y) { x = x + y >= mod ? x + y - mod : x + y; }
int mul (int x, int y) { return 1ll * x * y % mod; }

int n, q, A, mxs, all, rt, fa[_];
int a[_];
int sz[_], dis[_], dep[_], son[_], top[_], pa[_];
bool vis[_];
ll lst;

struct Info {
	int v;
	ll s;
	bool operator < (const Info & x) const { return v < x. v; }
} ;

vector <pair<int, int> > e[_];
vector <Info> v[2][_];

void dfs1 (int x, int fa) {
	sz[x] = 1;
	pa[x] = fa, dep[x] = dep[fa] + 1;
	for (auto & [y, w] : e[x]) {
		if (y == fa) continue ;
		dis[y] = dis[x] + w;
		dfs1(y, x), sz[x] += sz[y];
		if (sz[y] > sz[son[x]]) son[x] = y;
	}
}
void dfs2 (int x, int anc) {
	top[x] = anc;
	if (son[x]) dfs2(son[x], anc);
	for (auto & [y, w] : e[x]) if (y ^ pa[x] && y ^ son[x]) dfs2(y, y);
}
int dist (int x, int y) {
	int ret = dis[x] + dis[y], lc;
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) swap(x, y);
		x = pa[top[x]];
	}
	lc = dep[x] < dep[y] ? x : y;
	return ret - dis[lc] * 2;
}
void getrt (int x, int anc) {
	sz[x] = 1;
	int ms = 0;
	for (auto & [y, w] : e[x]) {
		if (vis[y] || y == anc) continue ;
		getrt(y, x), sz[x] += sz[y], ms = max(ms, sz[y]);
	}
	ms = max(all - sz[x], ms);
	if (ms < mxs) rt = x, mxs = ms;
}
void dfs3 (int x, int anc, int d) {
	sz[x] = 1;
	v[0][rt].push_back({a[x], d});
	if (fa[rt]) v[1][rt].push_back({a[x], dist(fa[rt], x)});
	for (auto & [y, w] : e[x]) {
		if (y == anc || vis[y]) continue ;
		dfs3(y, x, d + w);
		sz[x] += sz[y];
	}
}
void build (int x) {
	// cout << x << endl;
	vis[x] = 1, dfs3(x, 0, 0);
	for (auto & [y, w] : e[x]) {
		if (vis[y]) continue ;
		all = mxs = sz[y], rt = 0;
		getrt(y, x), fa[rt] = x;
		build(rt);
	}
}
ll query (int op, int x, int l, int r, ll &len) {
	int lp = lower_bound(v[op][x].begin(), v[op][x].end(), (Info){l, 0}) - v[op][x].begin() - 1,
		rp = upper_bound(v[op][x].begin(), v[op][x].end(), (Info){r, 0}) - v[op][x].begin() - 1;
	len = rp - lp;
	ll ret = 0;
	if (lp >= 0 && lp < v[op][x].size()) ret -= v[op][x][lp].s;
	if (rp >= 0 && rp < v[op][x].size()) ret += v[op][x][rp].s;
	return ret;
}
int main() {
	/*
	freopen(".in", "r", stdin);
	freopen(".out", "w", stdout);
	黛拉可玛莉·岗德森布莱德，一亿年一遇美少女。
	*/
	cin >> n >> q >> A;
	rep(i, 1, n) scanf("%d", & a[i]);
	rep(i, 1, n - 1) {
		int x, y, z;
		scanf("%d%d%d", & x, & y, & z);
		e[x].push_back({y, z}), e[y].push_back({x, z});
	}
	dfs1(1, 0), dfs2(1, 1);
	all = mxs = n, getrt(1, 0), build(rt);
	rep(x, 1, n) {
		sort(v[0][x].begin(), v[0][x].end()), 
		sort(v[1][x].begin(), v[1][x].end());
		for (int j = 1; j < v[0][x].size(); j ++)
			v[0][x][j].s += v[0][x][j - 1].s;
		for (int j = 1; j < v[1][x].size(); j ++)
		    v[1][x][j].s += v[1][x][j - 1].s;
	} 
	while (q --) {
		int x, l, r;
		ll p, q;
		scanf("%d%lld%lld", & x, & p, & q);
		l = min((p + lst) % A, (q + lst) % A),
		r = max((p + lst) % A, (q + lst) % A);
		ll t1, t2;
		lst = query(0, x, l, r, t1);
		for (int pos = x; fa[pos]; pos = fa[pos]) {
			lst += query(0, fa[pos], l, r, t1) - query(1, pos, l, r, t2);
			lst += (t1 - t2) * dist(fa[pos], x); 
		}
		printf("%lld\n", lst);
	}
	return 0;
}

刘伟在知乎上被找到帐号不是很正常。