傳送門
sb題。
直接用兩個 $multiset$

#include<bits/stdc++.h>
using namespace std;
inline int read(){
    int ans=0,w=1;
    char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();}
    while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar
 
();
    return ans*w;
}
const int N=5e5+5,inf=1e9+5e8;
int n,m,a[N],ans=inf,hd[N],tl[N];
multiset<int>T,R;
int main(){
    n=read(),m=read(),T.insert(-inf),T.insert(inf);
    for(int i=1;i<=n;++i){
        hd[i]=tl[i]=a[i]=read();
        T.insert(a[i]);
        multiset<int>::iterator it=
 
T.find(a[i]);
        --it,ans=min(ans,a[i]-(*it)),++it;
        ++it,ans=min(ans,(*it)-a[i]),--it;
    }
    for(int i=1;i<n;++i)R.insert(abs(a[i]-a[i+1]));
    while(m--){
        char s[10];
        int x,y;
        scanf("%s",s);
        if(s[0]=='I'){
            x=read(),y=read(),T.insert(y);
            multiset<int>::iterator it=T.find(y);
            --it,ans=min(ans,y-(*it)),++it;
            ++it,ans=min(ans,(*it)-y),--it; 
            if(x^n)R.erase(R.find(abs(hd[x+1]-tl[x]))),R.insert(abs(hd[x+1]-y));
            R.insert(abs(tl[x]-y)),tl[x]=y;
        }
        else if(s[4]=='G')printf("%d\n",*R.begin());
        else printf("%d\n",ans);
    }
    return 0;
}