https://codeforces.com/problemset/problem/1132/C
Idea: You can find n^2 enumeration.
For the state of the sequence after the first person is removed, O1 can be modified by difference. Then On accumulates.
How to calculate the answer of the sequence after removing the second person. It can be found that after the second person is removed, the state of the [l,r] interval left to be painted once is removed. So there is only one coverage for prefix and statistics.
Differential maintenance is good. Scan and process the prefix, and then On enumerate.
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=5e3+100;
typedef long long LL;
inline LL read(){LL x=0,f=1;char ch=getchar(); while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
struct P{
LL l,r;
}a[maxn];
LL cnt[maxn];///差分数组
LL pre[maxn];
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
LL n,q;cin>>n>>q;
for(LL i=1;i<=q;i++){
cin>>a[i].l>>a[i].r;
cnt[a[i].l]++;cnt[a[i].r+1]--;
}
LL ans=0;
for(LL i=1;i<=q;i++){///枚举去掉第一个人的序列状态
cnt[a[i].l]--;cnt[a[i].r+1]++;
LL tmp=0;LL res=0;
memset(pre,0,sizeof(pre));
for(LL j=1;j<=n;j++){
res+=cnt[j];
if(res>0) tmp++;
if(res==1) pre[j]=pre[j-1]+1;
else pre[j]=pre[j-1];
}
for(LL j=1;j<=q;j++){
if(i==j) continue;
ans=max(ans,tmp-(pre[a[j].r]-pre[a[j].l-1]));
}
cnt[a[i].l]++;cnt[a[i].r+1]--;
}
cout<<ans<<"\n";
return 0;
}