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【BZOJ 3672】【UOJ #7】【NOI 2014】购票

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$O(nlog^3n)$)或者点分治套cdq分治(时间复杂度$O(nlog^2n)$)。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$O(nlog^2n)$。

#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技术分享

【BZOJ 3672】【UOJ #7】【NOI 2014】购票