CF335F Buy One, Get One Free

\(\text{Solution}\)

其实不想写（因为不好写。。。
所以贴贴 \(\text{Solution}\)
然而就关于这题而言讲得也不太清楚
可撤销贪心就是维护当前状态的最优解，同时考虑以后的反悔
顺着题解，本题在 \(k<val\) 时必然换一种购买方式
而当 \(k\ge val\) 时，有两种方式，一种可能使当前解更优，但另一种却可以留下两次免费获取的机会
所以令 \(res=2val-k\)，压入堆中，对应当前解更优
而 \(res\) 以后可以被弹出，更改之前的决策，在当前免费取两次进而获得当前更优解

然而贴 \(\text{Code}\) 的原因是本人手打了个堆（几千年以来的壮举

\(\text{Code}\)

#include <cstdio>
#include <iostream>
#include <algorithm>
#define IN inline
using namespace std;
typedef long long ll;

const int N = 5e5 + 5;
int n, a[N], b[N], c[N], d[N];

IN void read(int &x) {
	x = 0; char ch = getchar(); int f = 1;
	for(; !isdigit(ch); f = (ch == '-' ? -1 : f), ch = getchar());
	for(; isdigit(ch); x = (x<<3)+(x<<1)+(ch^48), ch = getchar());
	x *= f;
}

struct Heap {
	int d[N], siz;
	IN Heap() {siz = 0;}
	IN void Up(int x) {while (x > 1 && d[x] < d[x >> 1]) swap(d[x], d[x >> 1]), x >>= 1;}
	IN void Down(int x) {
		int y = (x << 1);
		while ((y <= siz && d[x] > d[y]) || (y < siz && d[x] > d[y + 1])) {
			if (y < siz && d[y] > d[y + 1]) ++y;
			swap(d[x], d[y]), x = y, y = (x << 1);
		}
	}
	IN int size() {return siz;}
	IN int top() {return d[1];}
	IN void pop() {d[1] = d[siz--], Down(1);}
	IN void push(int x) {d[++siz] = x, Up(siz);}
}Q;

int main() {
	read(n);
	ll ans = 0;
	for(int i = 1; i <= n; i++) read(b[i]), ans += b[i];
	sort(b + 1, b + n + 1), reverse(b + 1, b + n + 1);
	int m = 0;
	for(int i = 1, j; i <= n; i = j + 1) {
		j = i;
		while (j < n && b[j + 1] == b[i]) ++j;
		a[++m] = b[i], c[m] = j - i + 1;
	}
	int num = 0;
	for(int i = 1; i <= m; num += c[i++]) {
		int rest = min(num - Q.size() * 2, c[i]), cnt = 0;
		for(int j = 1; j <= rest; j++) d[++cnt] = a[i];
		rest = c[i] - rest;
		for(int j = 1; j <= rest; j += 2) {
			if (!Q.size()) break;
			int k = Q.top(); Q.pop();
			if (k < a[i]) {d[++cnt] = a[i]; if (j < rest) d[++cnt] = a[i];}
			else {d[++cnt] = k; if (j < rest) d[++cnt] = a[i] * 2 - k;}
		}
		for(int j = 1; j <= cnt; j++) if (d[j] >= 0) Q.push(d[j]);
	}
	while (Q.size()) ans -= Q.top(), Q.pop();
	printf("%lld\n", ans);
}