// 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<bits/stdc++.h>
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<int, int>
#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<typename T> 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<pii> 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