luogu P2184 贪婪大陆

乍一不咋会

╭(╯3╰)╮

把地雷L到R看成一条线段

要求的就是区间内有多少条线段经过

很明显是要用[1,R]内的起点个数-[1,L-1]的终点个数

然后这起点和终点个数可以用简单的差分线段树来维护一下

其实树状数组更适合一些

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define ls rt<<1
#define rs rt<<1|1
using namespace std;
const int maxn = 1e5 + 7;
const int maxm = 4e5 + 7;

int n, m;
struct node {
	int l, r, size;
	int sum;
};
struct seg_tree {
	node e[maxm];
	void pushup(int rt) {
		e[rt].sum = e[ls].sum + e[rs].sum;
	}
	void build(int l, int r, int rt) {
		e[rt].l = l, e[rt].r = r;
		if (l == r) {
			return;
		}
		int mid = (l + r) >> 1;
		build(l, mid, ls);
		build(mid + 1, r, rs);
		pushup(rt);
	}

	void update(int L, int rt) {
		if (e[rt].l == e[rt].r) {
			e[rt].sum++;
			return;
		}
		int mid = (e[rt].l + e[rt].r) >> 1;
		if (L <= mid) update(L, ls);
		else update(L, rs);
		pushup(rt);
	}

	int query(int L, int R, int rt) {
		if (L <= e[rt].l && e[rt].r <= R) {
			return e[rt].sum;
		}
		int mid = (e[rt].l + e[rt].r) >> 1, ans = 0;
		if (L <= mid) ans += query(L, R, ls);
		if (R > mid) ans += query(L, R, rs);
		pushup(rt);
		return ans;
	}
} seg1, seg2;

int read() {
	int x = 0, f = 1; char s = getchar();
	for (; s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
	for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
	return x * f;
}

int main() {
	int n = read(), m = read();
	seg1.build(1, n, 1);
	seg2.build(1, n, 1);
	for (; m--;) {
		int tmp = read(), x = read(), y = read();
		if (tmp == 1) {
			seg1.update(x,1);
			seg2.update(y,1);
		} else {
			int ans = seg1.query(1,y,1) - seg2.query(1,x-1,1);
			printf("%d\n", ans);
		}
	}
	return 0;
}