P2801 教主的魔法 (线段树)

题目

P2801 教主的魔法

解析

成天做水题

线段树，第一问区间加很简单

第二问可以维护一个区间最大值和一个区间最小值，若C小于等于区间最小值，就加上区间长度，若C大于区间最大值，就加0

ps:求教指针线段树，我的空间怎么那么大

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
int n, m, num;
int a[N];
class tree {
	public :
		int mx, mn, sum;
		int lazy, len;
		tree *ls, *rs;
		tree() {
			sum = mx = mn = lazy = len = 0;
			ls = rs = NULL;
		}
} t[N];

template<class T>inline void read(T &x) {
	x = 0; int f = 0; char ch = getchar();
	while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	x = f ? -x : x;
	return;
}

inline void pushup(tree *rt) {
	rt->mn = min(rt->ls->mn, rt->rs->mn);
	rt->mx = max(rt->ls->mx, rt->rs->mx);
}

void build(int l, int r, tree *rt) {
	rt->len = r - l + 1;
	if (l == r) {
		rt->mn = rt->mx = a[l];
		return;
	}
	rt->ls = &t[++num], rt->rs = &t[++num];
	int m = (l + r) >> 1;
	build(l, m, rt->ls), build(m + 1, r, rt->rs);
	pushup(rt);
}

inline void pushdown(tree *rt) {
	if (rt->lazy) {
		rt->ls->mx += rt->lazy;
		rt->rs->mx += rt->lazy;
		rt->ls->mn += rt->lazy;
		rt->rs->mn += rt->lazy;
		rt->ls->lazy += rt->lazy;
		rt->rs->lazy += rt->lazy;
		rt->lazy = 0;
	}
}

void update(int L, int R, int c, int l, int r, tree *rt) {
	if (L <= l && r <= R) {
		rt->mn += c;
		rt->mx += c;
		rt->lazy += c;
		return;
	}
	pushdown(rt);
	int m = (l + r) >> 1;
	if (L <= m) update(L, R, c, l, m, rt->ls);
	if (R > m) update(L, R, c, m + 1, r, rt->rs);
	pushup(rt);
}

int query(int L, int R, int c, int l, int r, tree *rt) {
	if (L <= l && r <= R && rt->mn >= c) return rt->len;
	if (L <= l && r <= R && rt->mx < c) return 0;
	pushdown(rt);
	int m = (l + r) >> 1, ans = 0;
	if (L <= m) ans += query(L, R, c, l, m, rt->ls);
	if (R > m) ans += query(L, R, c, m + 1, r, rt->rs);
	return ans;
}

int main() {
	read(n), read(m);
	for (int i = 1; i <= n; ++i) read(a[i]);
	tree *rt = new tree;
	build(1, n, rt);
	for (int i = 1, x, y, z; i <= m; ++i) {
		char c;
		cin >> c;
		read(x), read(y), read(z);
		if (c == 'M')
			update(x, y, z, 1, n, rt);
		else
			printf("%d\n", query(x, y, z, 1, n, rt));
	}
	delete rt;
	return 0;
}