值周

良心出题人，给了512MB的空间，可以直接开下数组，差分就能A，不过都给到这么大了，差分就差点意思了，差分可以去试试校门外的树那道题，这里还是用离散去写吧，不知道为什么我用雨巨的离散这里只拿了90分……太奇怪了，好的我发现了是我太菜了看着思路写都有地方写错了，之前那个校门外的树都A过，那我还是按照我自己的思路来吧，计算被清除的区间长度，总长度减掉区间长度加一就是答案。维护终点即可。具体可以看下代码

#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;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();    return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e6 + 7;
pair<int, int> a[N]; //先first升序，后second升序

int main() {
    int n = read(), m = read();
    int l, r;
    for (int i = 1; i <= m; ++i)
        l = read(), r = read(), a[i] = make_pair(l, r);
    sort(a + 1, a + 1 + m);
    int sum = 0, ed = 0;
    for (int i = 1; i <= m; ++i)
        if (a[i].first >= ed) {
            sum += a[i].second - a[i].first + 1;
            ed = a[i].second;
        }
        else if (a[i].second >= ed) {
            sum += a[i].second - ed;
            ed = a[i].second;
        }
    write(n - sum + 1), putchar(10);
    return 0;
}