签到题模拟做法

和我一样一开始照着出题人描述补函数的举起爪子，我最后TLE吐了，才发现，如果按照他思路，时间爆炸了……
直接改掉函数，模拟就行了，其实也可以用线段树，当然更快，模拟更轻松一点就是了。）看来被set大常数卡的不轻，当然也感谢出题人仁慈没给极限数据……

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
typedef long long ll;
int a[maxn];
int opt, x, y;
set<pair<int, int> >P;
int main(){
    int n, L;
    cin >> n >> L;
    int ans = 0;
    while (n--){
        cin >> opt >> x >> y;
        if (opt == 1){
            if (P.find(make_pair(x, y)) != P.end()) continue;
            P.insert(make_pair(x, y));
            for (int i = x; i <= y; i++){
                a[i]++;
                if (a[i] == 1) ans++;
            }
        }
        else if (opt == 2){
            if (P.find(make_pair(x, y)) == P.end()) continue;
            P.erase(make_pair(x, y));
            for (int i = x; i <= y; i++){
                a[i]--;
                if (a[i] == 0) ans--;
            }
        }
        else{
            cout << ans << endl;
        }
    }
    return 0;
}

E正解、线段树写法

区间更新，区间查询，如果区间值大于1，正解求区间长度即可。否则就要去左右子树累加区间值大于1的区间长度，线段树操作

#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == &#39;-&#39;) w = -1; ch = getchar(); }    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();    return s * w; }

const int MAXN = 100010;
struct Node {
    int l, r;
    //ll lazy;
    ll sum;
} segTree[MAXN << 2];
void build(int i, int l, int r) {
    segTree[i].l = l;
    segTree[i].r = r;
    if (l == r) {
        //scanf("%lld", &segTree[i].sum);
        return;
    }
    int mid = l + r >> 1;
    build(i << 1, l, mid);
    build(i << 1 | 1, mid + 1, r);
    //segTree[i].sum = segTree[i << 1].sum + segTree[i << 1 | 1].sum;
}
void push_up(int i) {
    ll tmp = min(segTree[i << 1].sum, segTree[i << 1 | 1].sum);
    segTree[i].sum += tmp;
    segTree[i << 1].sum -= tmp;
    segTree[i << 1 | 1].sum -= tmp;
}
void push_down(int i) {
    if (segTree[i].sum) {
        segTree[i << 1].sum += segTree[i].sum;
        segTree[i << 1 | 1].sum += segTree[i].sum;
        segTree[i].sum = 0;
    }
}
void add(int i, int l, int r, ll v) {
    if (segTree[i].l >= l && segTree[i].r <= r) {
        segTree[i].sum += v;
        return;
    }
    push_down(i);
    int mid = segTree[i].l + segTree[i].r >> 1;
    if (mid >= l) add(i << 1, l, r, v);
    if (mid < r) add(i << 1 | 1, l, r, v);
    push_up(i);
}
ll query(int i, int l, int r) {
    ll res = 0;
    if (segTree[i].l >= l && segTree[i].r <= r)
        if (segTree[i].sum > 0)
            res += r - l + 1;
        else if (l != r) {
            int mid = segTree[i].l + segTree[i].r >> 1;
            push_down(i);
            res += query(i << 1, l, mid);
            res += query(i << 1 | 1, mid + 1, r);
            push_up(i);
        }
    return res;
}
int main() {
    int n, m, op;
    scanf("%d%d", &n, &m);
    build(1, 1, m);
    int l, r;
    while (n--) {
        op = read(), l = read(), r = read();
        if (op == 1)
            add(1, l, r, 1);
        else if (op == 2)
            add(1, l, r, -1);
        else
            printf("%lld\n", query(1, 1, m));
    }
    return 0;
}