noi.ac #38 线段树+时间复杂度分析

(des)
存在参数数组 (a)，(a) 升序排列

[a_1 < a_2 < cdots < a_m, m <= 10 ]

存在长度为 (n) 价值数组 (val)
存在 (3) 中操作

1. 使区间 ([l, r]) 内的 (val) 增加 (x)
2. 单点修改 (x)
3. 给定区间 ([l, r]) ，定义 (f(x)) 表示最大的 (i) 是的 (a_i <= x)
   求 (sum_{i = l} ^ {r} f(i))

(sol)
如果没有操作2，也就是说元素不会减小，同时 (f(x)) 也不会减小，所有的元素

(f(x)) 增加一共会有 (O(nm))。这里可以用线段树维护，第 (i) 个点维护的是

(f(i)) 还需要增加多少才可以增加，单次操作1相当于对区间 ([l, r]) 做减法

，显然如果某个时刻存在某个数 (<= 0)，这是 (f(x)) 需要增加，改变相关信息

，可以线段树维护区间最小值来实现。那么如果存在操作2是一样的，不过可能会

存在 (f(x)) 的减小的情况，并不会对时间复杂度产生大的影响
由于一共只会存在 (O((n + q)m)) 次增加，时间复杂度 O((n + 1)mlogn)。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>

using namespace std;
const int N = 1e5 + 10;

#define gc getchar()
#define Rep(i, a, b) for(int i = a; i <= b; i ++)
#define LL long long

inline int read() {int x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9')	x = x * 10 + c - '0', c = gc; return x;}
inline LL readLL() {LL x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}

int W[N << 2], F[N << 2], Minn[N << 2];
int A[15], Val[N];
int n, m, q;

#define lson jd << 1
#define rson jd << 1 | 1

void Build_tree(int l, int r, int jd) {
	if(l == r) {
		int x = lower_bound(A + 1, A + m + 1, Val[l]) - A;
		if(A[x] > Val[l]) x --;
		W[jd] = x;
		Minn[jd] = A[x + 1] - Val[l];
		return ;
	}
	int mid = (l + r) >> 1;
	Build_tree(l, mid, lson), Build_tree(mid + 1, r, rson);
	Minn[jd] = min(Minn[lson], Minn[rson]);
	W[jd] = W[lson] + W[rson];
}

void Push_down(int jd) {
	F[lson] += F[jd], F[rson] += F[jd];
	Minn[lson] += F[jd], Minn[rson] += F[jd];
	F[jd] = 0;
}

void Sec_G(int l, int r, int jd, int x, int y, int num) {
	if(x <= l && r <= y) {
		Minn[jd] -= num;
		F[jd] -= num;
		return ;
	}
	if(F[jd] != 0) Push_down(jd);
	int mid = (l + r) >> 1;
	if(x <= mid) Sec_G(l, mid, lson, x, y, num);
	if(y > mid)  Sec_G(mid + 1, r, rson, x, y, num);
	Minn[jd] = min(Minn[lson], Minn[rson]);
}

void Dfs_G(int l, int r, int jd) {
	if(l == r) {
		int b = A[W[jd] + 1] - Minn[jd];
		Val[l] = b;
		int x = lower_bound(A + 1, A + m + 1, b) - A;
		if(A[x] > b) x --;
		W[jd] = x;
		Minn[jd] = A[x + 1] - b;
		return ;
	}
	if(F[jd]) Push_down(jd);
	int mid = (l + r) >> 1;
	if(Minn[lson] <= 0) Dfs_G(l, mid, lson);
	if(Minn[rson] <= 0) Dfs_G(mid + 1, r, rson);
	Minn[jd] = min(Minn[lson], Minn[rson]);
	W[jd] = W[lson] + W[rson];
}

void Poi_G(int l, int r, int jd, int x, int num) {
	if(l == r) {
		Val[l] = num;
		int x = lower_bound(A + 1, A + m + 1, Val[l]) - A - 1;
		W[jd] = x;
		Minn[jd] = A[x + 1] - Val[l];
		return ;
	}
	if(F[jd]) Push_down(jd);
	int mid = (l + r) >> 1;
	if(x <= mid) Poi_G(l, mid, lson, x, num);
	else Poi_G(mid + 1, r, rson, x, num);
	W[jd] = W[lson] + W[rson];
	Minn[jd] = min(Minn[lson], Minn[rson]);
}

int Answer;

void Sec_A(int l, int r, int jd, int x, int y) {
	if(x <= l && r <= y) {
		Answer += W[jd];
		return ;
	}
	if(F[jd]) Push_down(jd);
	int mid = (l + r) >> 1;
	if(x <= mid) Sec_A(l, mid, lson, x, y);
	if(y > mid)  Sec_A(mid + 1, r, rson, x, y);
}

int main() {
	n = read(), m = read(), q = read();
	Rep(i, 1, m) A[i] = read();
	A[m + 1] = (1 << 30);
	Rep(i, 1, n) Val[i] = read();
	Build_tree(1, n, 1);
	Rep(t, 1, q) { 
		int opt = read();
		if(opt == 1) {
			int l = read(), r = read(), x = read();
			Sec_G(1, n, 1, l, r, x);
			if(Minn[1] <= 0) Dfs_G(1, n, 1);
		} else if(opt == 2) {
			int p = read(), x = read();
			Poi_G(1, n, 1, p, x);
		} else {
			int x = read(), y = read();
			Answer = 0;
			Sec_A(1, n, 1, x, y);
			cout << Answer << "
";
		}
	}
	return 0;
}