题目链接

思路

set+map+优先队列就可以水过去。可以发现，每插入一个元素，都会使得操作2中原来相邻的那个差值消失，然后多了两个新的差值。对于新的差值，只要直接扔到优先队列里就好了。那么删除呢。可以用map记录一下当前元素被删除了多少次。然后查询的时候将被删除的跳过即可。对于操作3，只要将插入的数都扔到set里，然后每次插入都找出当前点的前驱后继，并用ans记录下最优的。查询的时候直接输出ans即可。
为了证明自己真的有在练习平衡树，写了个splay代替set，然后跑的比set还慢2333。还可以再写一棵平衡树，来代替map+优先队列的操作。

STL代码

/*
* @Author: wxyww
* @Date:   2018-12-09 19:19:23
* @Last Modified time: 2018-12-09 19:56:45
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<set>
#include<cstring>
#include<cmath>
#include<map>
#include<ctime>
#include<queue>
#include<bitset>
using namespace std;
typedef long long ll;
const int N = 1000000 + 100,INF = 1e9;
map<int,int>ma;
priority_queue<int,vector<int>,greater<int> >q;
set<int>s;
ll read() {
   ll x=0,f=1;char c=getchar();
   while(c<'0'||c>'9') {
      if(c=='-') f=-1;
      c=getchar();
   }
   while(c>='0'&&c<='9') {
      x=x*10+c-'0';
      c=getchar();
   }
   return x*f;
}
int a[N],lianbiao[N],now[N];
char c[20];
set<int>::iterator k1,k2;
int ans1 = INF,ans2 = INF;
int main() {
   int n = read(),m = read();
   memset(a,-0x3f,sizeof(a));
   s.insert(-INF);s.insert(INF);
   for(int i = 1;i <= n;++i) {
      a[i] = read();
      lianbiao[i - 1] = i;
      k1 = --s.lower_bound(a[i]);k2 = s.lower_bound(a[i]);

      ans2 = min(ans2,min(abs(*k1 - a[i]),abs(*k2 - a[i])));
      q.push(abs(a[i] - a[i - 1]));
      s.insert(a[i]);
      now[i] = i;
   }
   while(m--) {
      scanf("%s",c + 1);
      if(c[1] == 'I') {
         int pos = read();
         a[++n] = read();
         int k = now[pos];
         
         ma[abs(a[k] - a[lianbiao[k]])]++;
         q.push(abs(a[n] - a[k]));q.push(abs(a[n] - a[lianbiao[k]]));

         k1 = --s.lower_bound(a[n]);k2 = s.lower_bound(a[n]);
         ans2 = min(ans2,min(abs(*k1 - a[n]),abs(*k2 - a[n])));
         s.insert(a[n]);

         now[pos] = n;
         lianbiao[n] = lianbiao[k];
         lianbiao[k] = n;
      }
      else if(c[6] == 'A') {
         int z = q.top();
         while(ma[z]) {
            q.pop();
            ma[z]--;
            z = q.top();
         }
         printf("%d\n",z);
      }
      else printf("%d\n",ans2);
   }
   return 0;
}

splay代码

/*
* @Author: wxyww
* @Date:   2018-12-09 16:29:51
* @Last Modified time: 2018-12-09 20:32:00
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<map>
#include<queue>
#include<ctime>
#include<bitset>
using namespace std;
#define ls TR[cur].ch[0]
#define rs TR[cur].ch[1]
typedef long long ll;
const int N = 1000000 + 100,INF = 1e9;
ll read() {
   ll x=0,f=1;char c=getchar();
   while(c<'0'||c>'9') {
      if(c=='-') f=-1;
      c=getchar();
   }
   while(c>='0'&&c<='9') {
      x=x*10+c-'0';
      c=getchar();
   }
   return x*f;
}
struct node {
   int ch[2],val,siz,pre,cnt;
}TR[N];
inline void up(int cur) {
   TR[cur].siz = TR[ls].siz + TR[rs].siz + TR[cur].cnt;
}
int tot,rt;
inline int getwh(int cur) {
   return TR[TR[cur].pre].ch[1] == cur;
}
inline void rotate(int cur) {
   int f = getwh(cur),fa = TR[cur].pre,gr = TR[fa].pre;
   TR[cur].pre = gr;
   TR[gr].ch[getwh(fa)] = cur;
   TR[TR[cur].ch[f ^ 1]].pre = fa;
   TR[fa].ch[f] = TR[cur].ch[f ^ 1];
   TR[fa].pre = cur;
   TR[cur].ch[f ^ 1] = fa;
   up(fa);up(cur);
}
void splay(int cur,int to) {
   while(TR[cur].pre != to) {
      if(TR[TR[cur].pre].pre != to) {
         if(getwh(cur) == getwh(TR[cur].pre)) rotate(TR[cur].pre);
         else rotate(cur);
      }
      rotate(cur);
   }
   if(!to) rt = cur;
}
void insert(int cur,int val,int father) {
   if(!cur) {
      cur = ++tot;
      TR[cur].pre = father;
      TR[cur].val = val;
      TR[cur].siz = TR[cur].cnt = 1;
      TR[father].ch[val > TR[father].val] = cur;
      splay(cur,0);
      return;
   }
   TR[cur].siz++;
   if(val == TR[cur].val) {TR[cur].cnt++;return;}
   insert(TR[cur].ch[val > TR[cur].val],val,cur);
}
inline int getpos(int val) {
   int cur = rt,lst;
   while(cur) {
      lst = cur;
      if(TR[cur].val == val) return cur;
      if(val > TR[cur].val) cur = rs;
      if(val < TR[cur].val) cur = ls;
   }
   return lst;
}
inline int pred(int val) {
   int cur = getpos(val);
   if(TR[cur].val <= val) return TR[cur].val;
   splay(cur,0);
   cur = ls;
   while(rs) cur = rs;
   return TR[cur].val;
}
inline int nex(int val) {
   int cur = getpos(val);
   if(TR[cur].val >= val) return TR[cur].val;
   splay(cur,0);
   cur = rs;
   while(ls) cur = ls;
   return TR[cur].val;
}
int a[N],lianbiao[N],now[N];
char c[20];
int ans1 = INF,ans2 = INF;
map<int,int>ma;
priority_queue<int,vector<int>,greater<int> >q;
int main() {

   int n = read(),m = read();
   memset(a,-0x3f,sizeof(a));
   insert(rt,-INF,0);insert(rt,INF,0);
   for(int i = 1;i <= n;++i) {
      a[i] = read();
      lianbiao[i - 1] = i;
      // puts("!!");
      int k1 = pred(a[i]),k2 = nex(a[i]);
      ans2 = min(ans2,min(abs(k1 - a[i]),abs(k2 - a[i])));
      q.push(abs(a[i] - a[i - 1]));
      insert(rt,a[i],0);
      now[i] = i;
   }
   while(m--) {
      scanf("%s",c + 1);
      if(c[1] == 'I') {
         int pos = read();
         a[++n] = read();
         int k = now[pos];
         
         ma[abs(a[k] - a[lianbiao[k]])]++;
         q.push(abs(a[n] - a[k]));q.push(abs(a[n] - a[lianbiao[k]]));

         int k1 = pred(a[n]),k2 = nex(a[n]);
         ans2 = min(ans2,min(abs(k1 - a[n]),abs(k2 - a[n])));
         insert(rt,a[n],0);

         now[pos] = n;
         lianbiao[n] = lianbiao[k];
         lianbiao[k] = n;
      }
      else if(c[6] == 'A') {
         int z = q.top();
         while(ma[z]) {
            q.pop();
            ma[z]--;
            z = q.top();
         }
         printf("%d\n",z);
      }
      else printf("%d\n",ans2);
   }
   return 0;
}

一言

积极者相信只有推动自己才能推动世界，只要推动自己就能推动世界。