[Ynoi2013] D2T2 做题记录

link

很厉害的题目。

先弱化题目，考虑 \(l = 1, r = n\) 怎么做。

设 \(f(x, y)\) 表示只保留值域在 \([a_x, a_y]\) 的数的最大子段和，有用状态数为 \(\mathcal O(n^2)\)。

我们发现这玩意其实是可以分治的：将原序列分成两半，求出左半部分和右半部分的只保留 \([a_x, a_y]\) 中的数的总和、最大前缀和、最大后缀和、最大子段和，合并是容易的。

这样时间复杂度为 \(T(n) = \mathcal O(n ^ 2) + 2T(\frac n2) = \mathcal O(n ^ 2)\)。

为了平衡复杂度，考虑分块，对于每块分治求出所有 \(f(x, y)\)，然后每个询问再依次合并每个块的答案。

这十分容易拓展到 \([l, r]\) 一般的情况，时间复杂度 \(\mathcal O(n\sqrt n)\)。

• 启示：弱化问题；分块平衡复杂度（不弱化不能从这个点入手）；无脑尝试分块；半群的分治性质

点击查看代码

#include <bits/stdc++.h>

namespace Initial {
	#define ll int
	#define ull unsigned long long
	#define fi first
	#define se second
	#define mkp make_pair
	#define pir pair <ll, ll>
	#define pb emplace_back
	#define i128 __int128
	using namespace std;
	const ll maxn = 1e5 + 10, inf = 1e9, mod = 1e9 + 7;
	ll power(ll a, ll b = mod - 2) {
		ll s = 1;
		while(b) {
			if(b & 1) s = 1ll * s * a %mod;
			a = 1ll * a * a %mod, b >>= 1;
		} return s;
	}
	template <class T>
	const ll pls(const T x, const T y) { return x + y >= mod? x + y - mod : x + y; }
	template <class T>
	void add(T &x, const T y) { x = x + y >= mod? x + y - mod : x + y; }
	template <class T>
	void chkmax(T &x, const T y) { x = x < y? y : x; }
	template <class T>
	void chkmin(T &x, const T y) { x = x > y? y : x; }
} using namespace Initial;

namespace Read {
	char buf[1 << 22], *p1, *p2;
	#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, (1 << 22) - 10, stdin), p1 == p2)? EOF : *p1++)
	template <class T>
	void rd(T &x) {
		char ch; bool neg = 0;
		while(!isdigit(ch = getchar()))
			if(ch == '-') neg = 1;
		x = ch - '0';
		while(isdigit(ch = getchar()))
			x = (x << 1) + (x << 3) + ch - '0';
		if(neg) x = -x;
	}
} using Read::rd;

namespace Fwrite {
	const int BUFFER_SIZE = 1 << 22, N = 15 + 7;
	char buffer[BUFFER_SIZE + 7], temp[N];
	char *p = buffer;
	
	inline void flush(){
		fwrite(buffer, p - buffer, 1, stdout);
		p = buffer;
	}
	
	inline void putchar(char ch){
		if (p - buffer >= BUFFER_SIZE) flush();
		*p++ = ch;
	}
	
	inline void write(long long x){
		int top = 0;
		do {
			temp[top++] = x % 10 + '0';
			x /= 10;
		} while (x > 0);
		while (top-- > 0) putchar(temp[top]);
	}
}

const ll B = 220;
ll n, m, a[maxn], ql[maxn], qr[maxn], qL[maxn], qR[maxn];
ll b[B + 10], id[B + 10], _id[B + 10], h[maxn], ht, _a[maxn];

struct Data { long long res, pre, suf, sum; }
 d[B + 10][B + 10], _d[B + 10][B + 10], ans[maxn];
const Data operator + (const Data a, const Data b) {
	return (Data) { max(max(a.res, b.res), a.suf + b.pre),
	 max(a.pre, a.sum + b.pre), max(b.suf, a.suf + b.sum), a.sum + b.sum };
}

void solve(const ll l, const ll r) {
	if(l == r) {
		ll w = max(0, b[l]);
		d[l][l] = (Data) { w, w, w, b[l] }; return;
	} const ll mid = l + r >> 1;
	solve(l, mid), solve(mid + 1, r);
	ll o1 = l, o2 = mid + 1, o = l;
	while(o1 <= mid || o2 <= r) {
		if(o1 <= mid && (o2 > r || b[id[o1]] <= b[id[o2]]))
			_id[o++] = id[o1++];
		else _id[o++] = id[o2++];
	}
	for(register ll i = l; i <= r; i = -~i) id[i] = _id[i];
	for(register ll i = l; i <= r; i = -~i) {
		const ll c = id[i]; auto p = _d[c];
		ll l_st = 0, l_ed = 0, r_st = 0, r_ed = 0;
		for(ll j = i; j <= r; j = -~j) {
			const ll x = id[j];
			if(x <= mid)  l_ed = x, !l_st && (l_st = x);
			else  r_ed = x, !r_st && (r_st = x);
			if(l_st && r_st)
				p[x] = d[l_st][l_ed] + d[r_st][r_ed];
			else p[x] = l_st? d[l_st][l_ed] : d[r_st][r_ed];
		}
	}
	for(register ll i = l; i <= r; i = -~i)
		for(register ll j = l; j <= r; j = -~j) d[i][j] = _d[i][j];
}
ll f[maxn], g[maxn];

int main() {
//	freopen("p.in", "r", stdin);
//	freopen("p.out", "w", stdout);
	rd(n), rd(m);
	for(register ll i = 1; i <= n; i = -~i) rd(a[i]), h[++ht] = a[i];
	sort(h + 1, h + 1 + ht);
	ht = unique(h + 1, h + 1 + ht) - h - 1;
	for(register ll i = 1; i <= m; i = -~i) {
		rd(ql[i]), rd(qr[i]), rd(qL[i]), rd(qR[i]);
		qL[i] = lower_bound(h + 1, h + 1 + ht, qL[i]) - h;
		qR[i] = upper_bound(h + 1, h + 1 + ht, qR[i]) - h - 1;
	}
	for(register ll i = 1; i <= n; i = -~i)
		_a[i] = lower_bound(h + 1, h + 1 + ht, a[i]) - h;
	for(register ll u = 1; (u - 1) * B < n; u = -~u) {
//		if(2 * u * B > n) return 0;
		const ll ul = (u - 1) * B + 1, ur = min(n, u * B);
		for(ll i = 0; i <= ur - ul; i++)
			b[i + 1] = a[ul + i], id[i + 1] = i + 1;
		solve(1, ur - ul + 1);
		memset(f, 0, sizeof f), memset(g, 0, sizeof g);
		for(register ll i = 1; i <= ur - ul + 1; i = -~i)
			f[_a[ul - 1 + id[i]]] = id[i];
		for(ll i = ur - ul + 1; i; i--)
			g[_a[ul - 1 + id[i]]] = id[i];
		for(register ll i = 1; i <= ht; i = -~i) !f[i] && (f[i] = f[i - 1]);
		for(register ll i = ht; i; i--) !g[i] && (g[i] = g[i + 1]);
		for(register ll i = 1; i <= m; i = -~i)
			if(ql[i] <= ul && ur <= qr[i]) {
				ll x = g[qL[i]], y = f[qR[i]];
				if(b[x] <= b[y])
					ans[i] = ans[i] + d[x][y];
			}
			else {
				ll x = max(ql[i], ul), y = min(qr[i], ur);
				if(x > y) continue;
				for(ll j = x; j <= y; j = -~j)
					if(_a[j] >= qL[i] && _a[j] <= qR[i]) {
						ans[i].suf = max(ans[i].suf + a[j], 0ll);
						ans[i].res = max(ans[i].res, ans[i].suf);
						ans[i].sum += a[j];
						ans[i].pre = max(ans[i].pre, ans[i].sum);
					}
			}
	}
	for(ll i = 1; i <= m; i++) Fwrite::write(ans[i].res), Fwrite::putchar('\n');
	Fwrite::flush();
	return 0;
}