暴力数据结构-珂朵莉树

珂朵莉树模板

一种使用std：set维护区间问题的暴力数据结构， 当数据足够随机时有良好的表现

缺点： 可能会被特殊数据卡成暴力的复杂度

参考博客

基本操作 ：split() //返回指定位置的迭代器
assign() 区间推平

板子：

struct Node{   //珂朵莉树基本结构
    int l, r;
    mutable LL val;
    
    Node(int l_ = -1, int r_ = -1, LL val_ = 0){
        l = l_; r = r_; val = val_;
    }
    
    bool operator < (const Node& t) const{
        return l < t.l;
    }

}; 

IT split(int  pos){   //split操作
    IT it = St.lower_bound(Node(pos));
    if(it!=St.end() && it->l == pos) return it;
    it--;
    Node tmp = *it;
    St.erase(it);
    St.insert(Node(tmp.l, pos - 1, tmp.val));
    return St.insert(Node(pos, tmp.r, tmp.val)).first;
    
}


void assign(int l, int r, LL val){  //区间推平操作
    IT itr = split(r + 1), itl = split(l);
    St.erase(itl, itr);
    St.insert(Node(l, r, val));

}

模板题

#include<bits/stdc++.h>
using namespace std;
#define mod %
typedef long long LL;

LL a[100010];

int n, m; LL seed, vmax;
struct Node{
    int l, r;
    mutable LL val;
    
    Node(int l_ = -1, int r_ = -1, LL val_ = 0){
        l = l_; r = r_; val = val_;
    }
    
    bool operator < (const Node& t) const{
        return l < t.l;
    }

};

set<Node> St;
typedef set<Node>::iterator IT;

IT split(int  pos){
    IT it = St.lower_bound(Node(pos));
    if(it!=St.end() && it->l == pos) return it;
    it--;
    Node tmp = *it;
    St.erase(it);
    St.insert(Node(tmp.l, pos - 1, tmp.val));
    return St.insert(Node(pos, tmp.r, tmp.val)).first;
    
}

void assign(int l, int r, LL val){
    IT itr = split(r + 1), itl = split(l);
    St.erase(itl, itr);
    St.insert(Node(l, r, val));

}

void add(int l, int  r, LL val){
    IT itr =  split(r + 1), itl = split(l);
    
    for(IT it = itl; it != itr; it++) it->val += val;
    
}
LL ksm(LL a, LL b, LL m){
	LL res = 1;
	while(b){
		if(b & 1) res = res * a % m;
		b >>= 1;
		a = a * a % m;

	}
	return res;

}
LL query(int l, int r, LL x, LL y){
    IT itr =  split(r + 1), itl =  split(l);
    LL res = 0;
    for(IT it = itl; it != itr; it++) res = (res  + ksm(it->val % y, x, y) % y * (it->r - it->l + 1) % y )% y;
    return res ;
    
}
LL findkth(int l, int r, int k){
    vector<pair<LL, int>> v;
    IT itr = split(r + 1), itl = split(l);
     for(IT it = itl; it != itr; it++) v.push_back({it->val, it->r - it-> l+ 1});
     	sort(v.begin(), v.end());

     for(auto item : v){
         k -= item.second;
         if(k <= 0) return item.first;
          
     }
     return 0;
     
}

LL rnd(){
    LL ret = seed;
    seed = (seed * 7 + 13) % 1000000007;
    return ret;
}

int main(){
    cin >> n >> m >> seed >> vmax;
    
    for(int i = 1; i <= n; i++){
        a[i] = (rnd() % vmax) + 1; 
    }
    // for(int i = 1; i <= n; i++) cout << a[i] << ' ';
    // cout << endl;
   int L, R, V;
   L = R = -1;
   for(int i = 1; i <= n; i++){
       if(L == -1 && R == -1) {
          L = i, R = i, V = a[i];
       }
       if(a[i] == V) R = i;
       else{
           St.insert(Node(L, R, V));
        L = R = i;
        V = a[i];
        
       }
   }
   St.insert(Node(L, R, V));

    LL op, l, r, x, y;

    for(int i = 1; i <= m; i++){
        
    op = (rnd() mod 4) + 1;
    l = (rnd() mod n) + 1;
    r = (rnd() mod n) + 1;

    if (l > r)
         swap(l, r);

    if (op == 3)
        x = (rnd() mod (r - l + 1)) + 1;
    else
        x = (rnd() mod vmax) + 1;

    if (op == 4)
        y = (rnd() mod vmax) + 1;
  
    
    if(op == 1) add(l, r, x);
    else if(op == 2) assign(l, r, x);
     else   	if(op == 3) cout << findkth(l, r, x)  << '\n';
     else  cout << query(l, r, x, y) << '\n'; 
   
    }
 
}