• https://darkbzoj.cc/problem/2006
• https://www.luogu.com.cn/problem/P2048

可以先写一个暴力 RMQ 解决 K=1 的 10 分代码，确保自己没读错题理解成字符串问题了囧。
那么 top K 可以用类似 [USACO3.1]丑数 Humble Numbers 那个题里的方法，开个堆，每次找到一个数就分裂一下塞回去。

这个题数据量足够大，询问数组的下标从 0 开始，还要询问 rmq 的位置，非常适合用我们新学习的 O(n)-O(1) RMQ 大显身手！。。。

#include <lastweapon/bitwise>

using namespace lastweapon;

const int N = int(5e5) + 9;
int s[N];
int n, K, L, R;

template<typename T> struct rmq {
	vector<T> v; int n;
	static const int b = 30; // block size
	vector<int> mask, t; // mask and sparse table

	int op(int x, int y) {
		return v[x] < v[y] ? x : y;
	}
	// least significant set bit
	int lsb(int x) {
		return x & -x;
	}
	// index of the most significant set bit
	int msb_index(int x) {
		return __builtin_clz(1)-__builtin_clz(x);
	}
	// answer query of v[r-size+1..r] using the masks, given size <= b
	int small(int r, int size = b) {
		// get only 'size' least significant bits of the mask
		// and then get the index of the msb of that
		int dist_from_r = msb_index(mask[r] & ((1<<size)-1));

		return r - dist_from_r;
	}

	rmq(){}

	rmq(const vector<T>& v_) : v(v_), n(v.size()), mask(n), t(n) {
		int curr_mask = 0;
		for (int i = 0; i < n; i++) {

			// shift mask by 1, keeping only the 'b' least significant bits
			curr_mask = (curr_mask<<1) & ((1<<b)-1);

			while (curr_mask > 0 and op(i, i - msb_index(lsb(curr_mask))) == i) {
				// current value is smaller than the value represented by the
				// last 1 in curr_mask, so we need to turn off that bit
				curr_mask ^= lsb(curr_mask);
			}
			// append extra 1 to the mask
			curr_mask |= 1;

			mask[i] = curr_mask;
		}

		// build sparse table over the n/b blocks
		// the sparse table is linearized, so what would be at
		// table[j][i] is stored in table[(n/b)*j + i]
		for (int i = 0; i < n/b; i++) t[i] = small(b*i+b-1);
		for (int j = 1; (1<<j) <= n/b; j++) for (int i = 0; i+(1<<j) <= n/b; i++)
			t[n/b*j+i] = op(t[n/b*(j-1)+i], t[n/b*(j-1)+i+(1<<(j-1))]);
	}
	// query(l, r) returns the actual minimum of v[l..r]
	// to get the index, just change the first and last lines of the function
	T query(int l, int r) {
		// query too small
		//if (r-l+1 <= b) return v[small(r, r-l+1)];
		if (r-l+1 <= b) return small(r, r-l+1);

		// get the minimum of the endpoints
		// (there is no problem if the ranges overlap with the sparse table query)
		int ans = op(small(l+b-1), small(r));

		// 'x' and 'y' are the blocks we need to query over
		int x = l/b+1, y = r/b-1;

		if (x <= y) {
			int j = msb_index(y-x+1);
			ans = op(ans, op(t[n/b*j+x], t[n/b*j+y-(1<<j)+1]));
		}

		//return v[ans];
		return ans;
	}
};


rmq<int> T;

struct rec {
    int i, l, r, m, f;
    rec(int i, int l, int r) : i(i), l(l), r(r), m(T.query(l, r)) {
        f = s[i] - s[m];
    }
    bool operator < (const rec& r) const {
        return f < r.f;
    }
};

int main() {

#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("/Users/minakokojima/Documents/GitHub/ACM-Training/Workspace/out.txt", "w", stdout);
#endif

    RD(n, K, L, R);

    VI a; a.PB(0);
    REP_1(i, n) RD(s[i]) += s[i-1], a.PB(s[i]); a.pop_back();

    T = rmq<int>(a);

    priority_queue<rec> Q;
    FOR_1(i, L, n) {
        Q.push(rec(i, max(0, i-R), i-L));
    }

    LL z = 0; DO(K) {
        auto u = Q.top(); Q.pop(); z += u.f;
        if (u.m != u.l) Q.push(rec(u.i, u.l, u.m-1));
        if (u.m != u.r) Q.push(rec(u.i, u.m+1, u.r));
    }

    cout << z << endl;
}

Posted by xiaodao
Category: 日常