P2048 [NOI2010] 超级钢琴 RMQ + Heap

妙妙妙。

我们可以考虑先求出一些区间的最优解 \((s, t, l, r, w)\) 表示 \([l, r]\) 的最优解为 \([s, t]\) 值为 \(w\).

再考虑将 \((s, maxpos(l, t - 1), l, r, sum[maxpos(l, t - 1)] - sum[s - 1])\) 和 \((s, maxpos(t + 1，ｒ), l, r, sum[maxpos(t + 1, r)] - sum[s - 1])\)　插入堆中, 最后取 \(k\) 个即可.

CODE :

// Cutslo 's CODE

// ST - RMQ

// 2022 - 3 - 31

#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#define int long long

using namespace std;

inline int read () {
    int ans = 0; char c = getchar(), last = ' ';
    while(c < '0' || c > '9') last = c,c = getchar();
    while(c >= '0' && c <= '9') ans = (ans << 1) + (ans << 3) + c - '0',c = getchar();
    return last == '-' ? - ans : ans;
}

const int N = 5e5 + 10;

int n, k, l, r, ans, sum[N], st[N][22];

struct NODE {
	int s, t, l, r;
	bool operator < (const NODE & x) const { return sum[t] - sum[s - 1] < sum[x.t] - sum[x.s - 1]; }
};
priority_queue <NODE> que;

inline void Init () {
	for (int i = 1; i <= n; i ++) st[i][0] = i;
	int t = (int)log2(n);
	for (int j = 1; j <= t; j ++) 
		for (int i = 1; i + (1 << j) <= n + 1; i ++) {
			int p = st[i][j - 1], q = st[i + (1 << (j - 1))][j - 1];
			st[i][j] = sum[p] > sum[q] ? p : q;
		}
}
inline int query (int l, int r) {
	int t = (int)log2(r - l + 1), p = st[l][t], q = st[r - (1 << t) + 1][t];
	return sum[p] > sum[q] ? p : q;
}

signed main () {
	n = read(), k = read(), l = read(), r = read();
	for (int i = 1; i <= n; i ++) sum[i] = read() + sum[i - 1];
	Init();
	for (int i = 1; i <= n; i ++) {
		int p = i + l - 1, q = min(i + r - 1, n);
		if (p <= n) que.push(NODE{i, query(p, q), p, q});
	}
	for (int i = 1; i <= k; i ++) {
		NODE st = que.top();
		int s = st.s, t = st.t, l = st.l, r = st.r;
		que.pop(), ans += sum[t] - sum[s - 1];
		if (l != t) que.push(NODE{s, query(l, t - 1), l, t - 1});
		if (r != t) que.push(NODE{s, query(t + 1, r), t + 1, r});
	}
	cout << ans;
	return 0;
}

启发很大。