abc343 F - Second Largest Query (线段树)

abc343 F - Second Largest Query (线段树)

题目链接

思路

看到题目发现显然是线段树的模板题目，难点是对于第二大数的个数该怎么维护，
线段树维护四个量：第一大数a1，第一大数的个数b1，第二大数a2，第二大数的个数b2，其他的代码套板子就行，主要是怎么pushup，可以对左右节点的第一第二大数排序，来得出父节点的第一第二大数，这个操作可以用map维护，较为简单，下面看代码

代码

#include<iostream>
#include<vector>
#include<map>
#define endl '\n'
using namespace std;
const int N = 2e5 + 10;
int n, q;
vector<int> nums(N + 1);

struct Node {
	int l, r;
	int a1, a2;
	int b1, b2;
}tr[N * 4];

void pushup(Node &t, Node l, Node r) {
	map<int, int> hask;//map来维护
	hask[l.a1] += l.b1, hask[l.a2] += l.b2, hask[r.a1] += r.b1, hask[r.a2] += r.b2;
	int cnt = 0;
	for (map<int, int>::reverse_iterator it = hask.rbegin(); it != hask.rend(); it++) {
		if (cnt == 0) t.a1 = it->first, t.b1 = it->second;
		if (cnt == 1) t.a2 = it->first, t.b2 = it->second;
		cnt++;
	}
}

void build(int u, int l, int r) {
	if (l == r) tr[u] = { l,r,nums[l],0,1,0 };
	else {
		tr[u] = { l,r };
		int mid = l + r >> 1;
		build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
		pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
	}
}

Node query(int u, int l, int r) {
	if (tr[u].l >= l && tr[u].r <= r) return tr[u];
	else {
		int mid = tr[u].l + tr[u].r >> 1;
		if (r <= mid) return query(u << 1, l, r);
		else if (l > mid) return query(u << 1 | 1, l, r);
		else {
			Node now;
			pushup(now, query(u << 1, l, r), query(u << 1 | 1, l, r));//合并答案
			return now;
		}
	}
}

void modify(int u, int p, int x) {
	if (tr[u].l == p && tr[u].r == p) tr[u].a1 = x;
	else {
		int mid = tr[u].l + tr[u].r >> 1;
		if (p <= mid) modify(u << 1, p, x);
		else modify(u << 1 | 1, p, x);
		pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
	}
}

void solve() {
	cin >> n >> q;
	for (int i = 1; i <= n; i++) cin >> nums[i];
	build(1, 1, n);
	while (q--) {
		int op, a, b;
		cin >> op >> a >> b;
		if (op == 1) modify(1, a, b);
		else cout << query(1, a, b).b2 << endl;
		/*Node now = query(1, a, b);
		cout << now.a1 << " " << now.b1 << " " << now.a2 << " " << now.b2 << endl;*/
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	solve();
	return 0;
}