Codeforces 1249 D2. Too Many Segments (hard version)

Portal

greedy

For the first illegal position, we obviously want to delete a few by covering its range to make the legal position

Obviously the right to delete point further to the right intervals is better, so we consider a priority right point right to delete, and then consider an illegal position

With a $ set $ maintenance at the right point and interval number can be

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
using namespace std;
typedef long long ll;
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') { x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
    return x*f;
}
const int N=2e5+7;
int n,m;
struct dat {
    int r,id;
    dat (int _r=0,int _id=0) { r=_r,id=_id; }
    inline bool operator < (const dat &tmp) const {
        return r!=tmp.r ? r<tmp.r : id<tmp.id;
    }
};
vector <dat> V[N];
vector <int> mov;
set <dat> S;
int main()
{
    n=read(),m=read(); int a,b,mx=0;
    for(int i=1;i<=n;i++)
        a=read(),b=read(),
        mx=max(mx,b),
        V[a].push_back(dat(b,i));
    int ans=0;
    for(int i=1;i<=mx;i++)
    {
        while(S.size()&&(*S.begin()).r<i) S.erase(S.begin());
        for(auto x: V[i]) S.insert(dat(x.r,x.id));
        while(S.size()>m)
        {
            auto p=S.rbegin();
            ans++,mov.push_back((*p).id);
            S.erase(*p);
        }
    }
    printf("%d\n",ans);
    for(auto x: mov) printf("%d ",x); puts("");
    return 0;
}