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;
}