CodeForces

Decription

区间最大字段和 (^+) ，最多选择 (k) 个子段，单点修改。

(n,qle 10^5,kle 20)

Solution

考虑暴力怎么做。

暴力大费流，拆点流量为 (1) 费用为 (a[i]) ，出点像下一个点的入点连边，增广 (k) 次并到 (costle 0) 为止。

考虑增广过程，其实就是每次取最大的一段，然后将该段去反（反向费用，退流）。

然后就可以线段树模拟了。

#include<bits/stdc++.h>
using namespace std;

template <class T> void read(T &x) {
	x = 0; bool flag = 0; char ch = getchar(); for (; ch < '0' || ch > '9'; ch = getchar()) flag |= ch == '-';
	for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48; flag ? x = ~x + 1 : 0;
}

//#pragma GCC diagnostic error "-std=c++14"

#define N 100010
#define rep(i, a, b) for (auto i = (a); i <= (b); ++i)
#define drp(i, a, b) for (auto i = (a); i >= (b); --i)
#define ll long long
#define INF 0x3f3f3f3f

struct Data {
	int sum, lval, rval, sval, sl, sr, lr, rl;
	Data(int _sum = 0, int _lval = 0, int _rval = 0, int _sval = 0, int _sl = 0, int _sr = 0, int _lr = 0, int _rl = 0) {
		sum = _sum, lval = _lval, rval = _rval, sval = _sval, sl = _sl, sr = _sr, lr = _lr, rl = _rl;
	}
	void reverse() {
		sum *= -1, lval *= -1, rval *= -1, sval *= -1;
	}
	void init(int t) {
		lval = rval = sval = INF;
		if (t) reverse();
	}
	void merge(const Data& l, const Data& r, bool t) {
		sum = l.sum + r.sum;
		if (t) {
			lval = max(l.lval, l.sum + r.lval), rval = max(r.rval, r.sum + l.rval);
			sval = max(max(l.sval, r.sval), l.rval + r.lval);
		}
		else {
			lval = min(l.lval, l.sum + r.lval), rval = min(r.rval, r.sum + l.rval);
			sval = min(min(l.sval, r.sval), l.rval + r.lval);
		}
		lr = (lval == l.lval ? l.lr : r.lr);
		rl = (rval == r.rval ? r.rl : l.rl);
		if (sval == l.sval) sl = l.sl, sr = l.sr;
		else if (sval == r.sval) sl = r.sl, sr = r.sr;
		else sl = l.rl, sr = r.lr;
	}
};

struct Node {
	Data mx, mn;
	void init() {
		mx.init(1), mn.init(0);
	}
}T[N << 2];

int n, a[N];
bool tag[N << 2];

#define ls rt << 1
#define rs ls | 1
#define mid (l + r >> 1)

void pushup(Node& now, const Node& l, const Node& r) {
	now.init();
	now.mx.merge(l.mx, r.mx, 1), now.mn.merge(l.mn, r.mn, 0);
}

void pushDown(int rt) {
	if (tag[rt]) {
		T[ls].mx.reverse(), T[rs].mx.reverse();
		T[ls].mn.reverse(), T[rs].mn.reverse();
		swap(T[ls].mx, T[ls].mn), swap(T[rs].mx, T[rs].mn);
		tag[ls] ^= 1, tag[rs] ^= 1, tag[rt] = 0;
	}
}

void update(int rt, int l, int r, int p) {
	if (l == r) {
		T[rt].mn = T[rt].mx = Data(a[l], a[l], a[l], a[l], l, r, l, r);
		return;
	}
	pushDown(rt);
	p <= mid ? update(ls, l, mid, p) : update(rs, mid + 1, r, p);
	pushup(T[rt], T[ls], T[rs]);
}

Node query(int rt, int l, int r, int L, int R) {
	if (L <= l && r <= R) return T[rt];
	pushDown(rt);
	Node ret, lret, rret;
	ret.init(), lret.init(), rret.init();
	if (L <= mid) lret = query(ls, l, mid, L, R);
	if (R > mid) rret = query(rs, mid + 1, r, L, R);
	if (R <= mid) ret = lret;
	else if (L > mid) ret = rret;
	else pushup(ret, lret, rret);
	return ret;
}

void update(int rt, int l, int r, int L, int R) {
	if (L <= l && r <= R) {
		T[rt].mx.reverse(), T[rt].mn.reverse();
		swap(T[rt].mx, T[rt].mn);
		tag[rt] ^= 1;
		return;
	}
	pushDown(rt);
	if (L <= mid) update(ls, l, mid, L, R);
	if (R > mid) update(rs, mid + 1, r, L, R);
	pushup(T[rt], T[ls], T[rs]);
}

int ans = 0;
void dfs(int k, int l, int r) {
	Node ret = query(1, 1, n, l, r);
	if (ret.mx.sval <= 0) return;
	ans += ret.mx.sval;
	update(1, 1, n, ret.mx.sl, ret.mx.sr);
	if (k - 1) dfs(k - 1, l, r);
	update(1, 1, n, ret.mx.sl, ret.mx.sr);
}

int main() {
	read(n);
	rep(i, 1, n) read(a[i]), update(1, 1, n, i);
	int q; read(q);
	while (q--) {
		int op; read(op);
		if (!op) {
			int p; read(p), read(a[p]);
			update(1, 1, n, p);
		}
		else {
			int l, r, k; read(l), read(r), read(k);
			ans = 0, dfs(k, l, r);
			printf("%d
", ans);
		}
	}
	return 0;
}