Codeforces 193D Two Segments (线段树)

最新推荐文章于 2023-09-12 10:51:00 发布

原创最新推荐文章于 2023-09-12 10:51:00 发布 · 2.5w 阅读

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本文介绍了一种使用线段树进行区间更新与查询的优化方法，通过递归地将操作应用于子节点并维护懒惰标记来避免重复计算。该方法特别适用于解决需要频繁更新区间内元素并查询区间最大最小值的问题。

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <ctime>
#include <cstdlib>
#include <functional>
#pragma comment(linker,"/STACK:102400000,102400000")
using namespace std;


#define eps 1e-10
#define N 300030
#define B 20
#define M 3000020
#define inf 0x3f3f3f3f
#define LL long long
#define pii pair<int, int>
#define MP make_pair
#define fi first
#define se second
#define mod 1000000007
#define ls (i << 1)
#define rs (ls | 1)
#define md (ll + rr >> 1)
#define lson ll, md, ls
#define rson md + 1, rr, rs


int mi1[N << 2], mi2[N << 2], s1[N << 2], s2[N << 2], lazy[N << 2];


int n, a[N], f[N];

void build(int ll, int rr, int i) {
	mi1[i] = 0, s1[i] = rr - ll + 1;
	mi2[i] = inf, s2[i] = 0;
	if(ll == rr) return;
	build(lson);
	build(rson);
}
LL ans;
int t;

void upd(int v, int i) {
	mi1[i] += v;
	mi2[i] += v;
	lazy[i] += v;
}

void push_down(int i) {
	if(lazy[i]) {
		upd(lazy[i], ls);
		upd(lazy[i], rs);
		lazy[i] = 0;
	}
}
void calc(int i, int x, int v) {
	if(v == 0) return;
	if(x < mi1[i]) {
		mi2[i] = mi1[i], s2[i] = s1[i];
		mi1[i] = x, s1[i] = v;
	}
	else if(x == mi1[i]) {
		s1[i] += v;
	}
	else if(x < mi2[i]) {
		mi2[i] = x, s2[i] = v;
	}
	else if(x == mi2[i]) {
		s2[i] += v;
	}
}
void push_up(int i) {
	mi1[i] = mi1[ls], s1[i] = s1[ls];
	mi2[i] = mi2[ls], s2[i] = s2[ls];
	calc(i, mi1[rs], s1[rs]);
	calc(i, mi2[rs], s2[rs]);
}
void update(int l, int r, int v, int ll, int rr, int i) {
	if(ll == l && rr == r) {
		upd(v, i);
		return;
	}
	push_down(i);
	if(r <= md) update(l, r, v, lson);
	else if(l > md) update(l, r, v, rson);
	else {
		update(l, md, v, lson);
		update(md + 1, r, v, rson);
	}
	push_up(i);
}

void query(int l, int r, int ll, int rr, int i) {
	if(ll == l && rr == r) {
		calc(t, mi1[i], s1[i]);
		calc(t, mi2[i], s2[i]);
		return;
	}
	push_down(i);
	if(r <= md) query(l, r, lson);
	else if(l > md) query(l, r, rson);
	else {
		query(l, md, lson);
		query(md + 1, r, rson);
	}
}
int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
		f[a[i]] = i;
	}
	build(1, n, 1);
	ans = 0;
	t = (N << 2) - 1;
	for(int i = 1; i <= n; ++i) {
		int l = -1, r = -1;
		if(f[i] > 1 && a[f[i] - 1] <= i) l = a[f[i] - 1];
		if(f[i] < n && a[f[i] + 1] <= i) r = a[f[i] + 1];
		if(l > r) swap(l, r);
		if(r == -1) update(1, i, 1, 1, n, 1);
		else if(l == -1) {
			update(r + 1, i, 1, 1, n, 1);
		}
		else {
			update(r + 1, i, 1, 1, n, 1);
			update(1, l, -1, 1, n, 1);
		}
		
		mi1[t] = inf, mi2[t] = inf;
		query(1, i, 1, n, 1);
		if(mi1[t] == 1) ans += s1[t];
		if(mi2[t] == 2) ans += s2[t];

	}
	printf("%lld\n", ans - n);
	return 0;
}