BZOJ 3672UOJ #7NOI 2014购票

Posted 2020-08-08 abclzr

http://www.lydsy.com/JudgeOnline/problem.php?id=3672

http://uoj.ac/problem/7

链上的情况可以用斜率优化dp。树上用斜率优化dp时，单调队列的复杂度是均摊$O(n)$的，所以放到树上做“可持久化单调队列”复杂度是$O(n^2)$的，所以不能树上斜率优化。

这道题可以用树链剖分(时间复杂度$O(n\\log^3n)$)或者点分治套cdq分治(时间复杂度$O(n\\log^2n)$)。因为树链剖分感觉比较难写，而且每个节点用vector存单调队列，显得比较卡空间，而且时间复杂度多一个log，所以写了点分治。

对于一个点$i$，从i到根的路径上有$j$，$k$。假设$k$的深度比$j$小，且用$k$来更新$i$比$j$更优，得出式子($dis$为到根的距离)：

$$\\frac{f_j-f_k}{dis_j-dis_k}>p_i$$

对于每个点，把$dis$看成横坐标，$f$看成纵坐标，最优点一定在下凸壳上。

对于一棵树，求出分治重心，再对重心上方的子树进行分治，分治完后重心到该树的根上所有的点的$f$值都求好了，然后就用重心到根这条链上所有的点去更新重心子树中所有的点。

先对重心子树中所有的点(不包括重心)按“$l$值-到重心的距离”排序，保证用来更新“子树中的点”的“重心到根上的点”到重心的距离单调递增，这样拿单调栈来维护一个下凸壳来更新就可以了。

时间复杂度$O(n\\log^2n)$。

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 200003;
int in() {
	int k = 0, fh = 1; char c = getchar();
	for(; c < \'0\' || c > \'9\'; c = getchar())
		if (c == \'-\') fh = -1;
	for(; c >= \'0\' && c <= \'9\'; c = getchar())
		k = (k << 3) + (k << 1) + c - \'0\';
	return k * fh;
}
ll inll() {
	ll k = 0; int fh = 1; char c = getchar();
	for(; c < \'0\' || c > \'9\'; c = getchar())
		if (c == \'-\') fh = -1;
	for(; c >= \'0\' && c <= \'9\'; c = getchar())
		k = (k << 3) + (k << 1) + c - \'0\';
	return k * fh;
}

bool vis[N];
struct node {
	int nxt, to; ll w;
	node(int _nxt = 0, int _to = 0, ll _w = 0) : nxt(_nxt), to(_to), w(_w) {}
} E[N << 1];
int t, n, fa[N], Q[N], cnt = 0, sz[N], point[N];
ll f[N], fadis[N], p[N], q[N], l[N], longdis[N];

void ins(int u, int v, ll w) {E[++cnt] = node(point[u], v, w); point[u] = cnt;}

int findrt(int x) {
	int u, head = 0, tail = 1; Q[1] = x;
	while (head != tail) {
		u = Q[++head]; sz[u] = 1;
		for(int i = point[u]; i; i = E[i].nxt)
			if (!vis[E[i].to] && E[i].to != fa[u])
				Q[++tail] = E[i].to;
	}
	for(int i = tail; i >= 1; --i) {
		if ((sz[Q[i]] << 1) > tail) return Q[i];
		sz[fa[Q[i]]] += sz[Q[i]];
	}
}

int tot, qu[N];
struct data {
	ll dis, line; int id;
	data(ll _dis = 0, ll _line = 0, int _id = 0) : dis(_dis), line(_line), id(_id) {}
	bool operator < (const data &A) const {
		return line < A.line;
	}
} a[N];

double k_num(int x, int y) {
	return 1.0 * (f[x] - f[y]) / (longdis[x] - longdis[y]);
}

int find(int le, double k) {
	int left = 0, right = le - 1, mid;
	while (left < right) {
		mid = (left + right) >> 1;
		if (k_num(qu[mid], qu[mid + 1]) > k) left = mid + 1;
		else right = mid;
	}
	if (left == le - 1 && k_num(qu[left], qu[le]) > k) return qu[le];
	return qu[left];
}

int cont = 0;
void cdq(int x) {
	vis[x] = true;
	int i, tmp, tail, head = 0, up = x; ll len = 0;
	for(tmp = point[x]; tmp; tmp = E[tmp].nxt)
		if (!vis[E[tmp].to] && E[tmp].to == fa[x]) {
			while (!vis[fa[up]] && up != 1) up = fa[up];
			cdq(findrt(up));
			break;
		}
	
	tot = 0;
	for(i = point[x]; i; i = E[i].nxt)
		if (!vis[E[i].to] && E[i].to != fa[x])
			a[++tot] = data(fadis[E[i].to], l[E[i].to] - fadis[E[i].to], E[i].to);
	
	while (head != tot) {
		++head;
		for(i = point[a[head].id]; i; i = E[i].nxt)
			if (!vis[E[i].to] && E[i].to != fa[a[head].id])
				a[++tot] = data(fadis[E[i].to] + a[head].dis, l[E[i].to] - fadis[E[i].to] - a[head].dis, E[i].to);
	}
	
	stable_sort(a + 1, a + tot + 1);
	
	tmp = x;
	while (tmp != up) {
		tmp = fa[tmp]; if (longdis[x] - longdis[tmp] > l[x]) break;
		if (f[x] == -1) f[x] = f[tmp] + (longdis[x] - longdis[tmp]) * p[x] + q[x];
		else f[x] = min(f[x], f[tmp] + (longdis[x] - longdis[tmp]) * p[x] + q[x]);
	}
	
	tail = 0; tmp = x; qu[0] = x;
	for(i = 1; i <= tot; ++i) {
		if (a[i].line < 0) continue;
		while (tmp != up && len + fadis[tmp] <= a[i].line) {
			len += fadis[tmp];
			tmp = fa[tmp];
			while (tail && k_num(tmp, qu[tail]) > k_num(qu[tail], qu[tail - 1]))
				--tail;
			qu[++tail] = tmp;
		}
		head = find(tail, (double) p[a[i].id]);
		if (f[a[i].id] == -1) f[a[i].id] = f[head] + (longdis[a[i].id] - longdis[head]) * p[a[i].id] + q[a[i].id];
		else f[a[i].id] = min(f[a[i].id], f[head] + (longdis[a[i].id] - longdis[head]) * p[a[i].id] + q[a[i].id]);
	}
	
	for(i = point[x]; i; i = E[i].nxt)
		if (!vis[E[i].to] && E[i].to != fa[x])
			cdq(findrt(E[i].to));
}

int main() {
	n = in(); t = in();
	for(int i = 2; i <= n; ++i) {
		fa[i] = in(); fadis[i] = inll();
		ins(fa[i], i, fadis[i]);
		ins(i, fa[i], fadis[i]);
		p[i] = inll(); q[i] = inll(); l[i] = inll();
	}
	
	memset(f, -1, sizeof(ll) * (n + 1));
	f[1] = 0;
	int u, head = 0, tail = 1;
	Q[1] = 1; longdis[1] = 0;
	while (head != tail) {
		u = Q[++head];
		for(int i = point[u]; i; i = E[i].nxt)
			if (E[i].to != fa[u]) {
				longdis[E[i].to] = longdis[u] + fadis[E[i].to];
				Q[++tail] = E[i].to;
			}
	}
	
	cdq(findrt(1));
	for(int i = 2; i <= n; ++i)
		printf("%lld\\n", f[i]);
	return 0;
}

终于调完了，写完后有好多错QAQ