二分+差分维护

金牌厨师
题意：
给定 $m$ 个区间，每个区间在 $[1,n]$ 之间，选出 $k$ 个区间，这 $k$ 个区间的交集长度为 $x$，定义满意度为 $min(k,x)$
求最大满意度
思路：
显然二分最大化
具体 $check$ 看代码好理解
非常妙的思路
code：

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define eps 1e-6
using namespace std;
const int maxn = 3e5 + 9;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll n, m;
/*
vector <int> R[maxn];
bool check(int x){
	priority_queue <int, vector<int>, greater<int> > q;
	for(int i = 1; i + x - 1 <= n; ++i){
		for(auto r : R[i]) q.push(r);
		while(!q.empty() && q.top() < i + x - 1) q.pop();
		if(q.size() >= x) return 1;
	}
	return 0;
}*/

pair <int, int> a[maxn];
int c[maxn];
bool check(int x){
	memset(c, 0, sizeof(c));
	for(int i = 1; i <= m; ++i){
		int r = a[i].first + x - 1;
		if(r <= a[i].second) c[r]++, c[a[i].second + 1] --;
	}
	for(int i = 1; i <= n; ++i) 
	{
		c[i] += c[i-1];
		if(c[i] >= x) return 1;
	}
	return 0;
}
void work()
{
	cin >> n >> m;
	/*for(int i = 1, l, r; i <= m; ++i){
		cin >> l >> r;
		R[l].push_back(r);
	}*/
	for(int i = 1; i <= m; ++i) cin >> a[i].first >> a[i].second;
	int l = 1, r = min(n, m);
	while(l < r)
	{
		int mid = (l + r + 1) >> 1;
		if(check(mid)) l = mid;
		else r = mid - 1;
	}
	cout << r;
}

int main()
{
	ios::sync_with_stdio(0);
//	int TT;cin>>TT;while(TT--)
	work();
	return 0;
}