Luogu4137 Rmq problem/mex 主席树

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用主席树水莫队题……

我们对于前缀和建立主席树，对于主席树中的每一个叶子节点表示它对应的数字最后出现的位置的编号，非叶子节点求左右节点的最小值，那么对于每一次询问$l,r$就是在第$r$棵主席树上找到权值$<l$的最左端的点，在主席树上二分即可。

 #include<bits/stdc++.h>
 #define mid ((l + r) >> 1)
 #define min(x,y) x < y ? x : y
 #define pushup(x) Tree[x].minN = min(Tree[Tree[x].l].minN , Tree[Tree[x].r].minN)
 //This code is written by Itst
 using namespace std;

 inline int read(){
     ;
     ;
     char c = getchar();
     while(c != EOF && !isdigit(c)){
         if(c == '-')
             f = ;
         c = getchar();
     }
     while(c != EOF && isdigit(c)){
         a = (a << ) + (a << ) + (c ^ ');
         c = getchar();
     }
     return f ? -a : a;
 }

 ;
 struct query{
     int l , r , ind;
 }que[MAXN];
 struct node{
     int l , r , minN;
 }Tree[MAXN * ];
 int ans[MAXN] , root[MAXN] , N , M , cntNode;

 void insert(int& x , int p , int l , int r , int tar , int num){
     x = ++cntNode;
     Tree[x] = Tree[p];
     if(l == r)
         Tree[x].minN = num;
     else{
         if(mid >= tar)
             insert(Tree[x].l , Tree[p].l , l , mid , tar , num);
         else
             insert(Tree[x].r , Tree[p].r , mid +  , r , tar , num);
         pushup(x);
     }
 }

 int query(int x , int l , int r , int minN){
     if(l == r)
         return l;
     if(Tree[Tree[x].l].minN < minN)
         return query(Tree[x].l , l , mid , minN);
      , r , minN);
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen("4137.in" , "r" , stdin);
     //freopen("4137.out" , "w" , stdout);
 #endif
     N = read();
     M = read();
      ; i <= N ; ++i){
         int a = read();
         if(a < N)
             insert(root[i] , root[i - ] ,  , N , a , i);
         else
             root[i] = root[i - ];
     }
      ; i <= M ; ++i){
         int a = read() , b = read();
         printf( , N , a));
     }
     ;
 }