ZOJ 3937 More Health Points （2016 浙江省赛 B题，可持久维护凸壳）

题目链接  2016 ZJCPC Problem B

题意  CF 660F的树上版本。

其他做的方法都差不多，关键是把凸壳放到树上。

每次确定扔掉几个元素的时候直接$O(1)$修改（先不清楚这个位置之后的元素因为之后还要恢复），然后$O(1)$恢复，通过这个来实现可持久。

#include <bits/stdc++.h>
 
using namespace std;
 
#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second
 
typedef long long LL;
 
const int N = 1e5 + 10;
 
int T;
int n;
int r;
int deep[N], q[N];
LL a[N], c[N], s[N];
LL ans;
vector <int> v[N];
 
inline LL y(int x){
	return c[x] - 1ll * deep[x] * s[x];
}
 
inline long double g(int j, int k){
	double dy = 1.00 * y(j) - 1.00 * y(k);
	double dx = 1.00 *  deep[j] - 1.00 * deep[k];
	return dy / dx;
}
 
inline LL calc(int x, int y){
	return c[x] - c[y] - 1ll * deep[y] * (s[x] - s[y]);
}
 
void pre(int x, int fa, int dep){
	deep[x] = dep;
	s[x] = s[fa] + a[x];
	c[x] = c[fa] + 1ll * dep * a[x];
 
	for (auto u : v[x]){
		pre(u, x, dep + 1);
	}
}
 
inline int pos(LL x, int tail){
	x = -x;
	int l = 1, r = tail - 1, ret = 0;
 
	while (l <= r){
		int mid = (l + r) >> 1;
		if (g(q[mid], q[mid - 1]) < x) l = (ret = mid) + 1;
		else r = mid - 1;
	}
 
	return q[ret];
}
 
inline int gettail(int x, int tail){
	int l = 2, r = tail, ret = 0;
 
	if (g(x, q[1]) < g(q[1], q[0])) return 1;
 
	while (l <= r){
		int mid = (l + r) >> 1;
		if (g(x, q[mid - 1]) >= g(q[mid - 1], q[mid - 2])) l = (ret = mid) + 1;
		else r = mid - 1;
	}
 
	return ret;
}
 
void dfs(int x, int tail){
 
	int y = pos(s[x], tail);
	ans = max(ans, calc(x, y));
 
	int cnt, t, re, la;
	if (tail <= 1 || g(x, q[tail - 1]) >= g(q[tail - 1], q[tail - 2])){
		re = tail;
		la = q[tail];
		q[tail] = x;
		tail++;
		cnt = tail;
	}	
 
	else{
		t = gettail(x, tail);    //get the position
		re = t;
		la = q[t];
		q[t] = x;
		cnt = t + 1;
	}
	//replace
 
	for (auto u : v[x]) dfs(u, cnt);  //continue solving
 
	q[re] = la;  //undo
}		
 
int main(){
 
	scanf("%d", &T);
	while (T--){
		scanf("%d", &n);
		rep(i, 0, n + 1) v[i].clear();
 
		rep(i, 1, n) scanf("%lld", a + i);
		rep(i, 2, n){
			int x;
			scanf("%d", &x);
			v[x].push_back(i);
		}
 
		pre(1, 0, 1);
 
		rep(i, 0, n + 1) q[i] = 0;
		r = 0;
		q[r++] = 0;
 
		ans = 0;
		dfs(1, r);
		printf("%lld\n", ans);
	}
 
	return 0;
}