华华听月月唱歌

// Problem: 华华听月月唱歌
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1046
// Memory Limit: 65536 MB
// Time Limit: 2000 ms
// 2022-02-25 19:00:33
// 
// Powered by CP Editor (https://cpeditor.org)

#include
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
bool cmp (pii a, pii b) {
    return a.fi < b.fi;
}
void solve() {
	int n, m;
    cin >> n >> m;
    vector range(m);
    for (int i = 0; i < m ;i ++) {
        int l, r;
        cin >> l >> r;
       range[i] = {l, r};
    }
    sort(all(range), cmp);
    int ans = 0;
    int r = 0, maxv = 0;
   for (int i = 0; i < m && r < n;) {
       ans ++;
       if (r + 1 < range[i].fi ) {
           cout <<-1 << "n";
           return;
       }
       while (i < m && range[i].fi <= r + 1)
       {
           maxv = max (maxv, range[i].se);
           i ++;
       }
       r = maxv; 
   }
    if (r >= n) cout << ans << endl;
    else cout << -1 << endl;
}
int main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

只需要记录右端点就行了。我做的时候想着记录两个端点 贪心策略找到最最大的右端点。注意这里<=r+1也是合法的。因为只需要包括点而不是区间