【OI】指针线段树&指针

对于线段树，我们一般需要n*4的空间去存储线段树，然后有一种玄学操作是用指针来实现线段树。

 

#include <inttypes.h>
#include <algorithm>
#include <cstdio>
#include <iostream>
#include <vector>
#define debug(x) std::cout<< #x << " = " << std::endl;

typedef long long int int_t;

using std::cin;
using std::cout;
using std::endl;

struct Node{
    
    Node *left,*right;
    int_t value;
    int begin,end;
    int mark;
    Node(int begin, int end) {
        this->begin = begin;
        this->end = end;
        left = right = NULL;
        mark = 0; 
    }
    
    void add(int_t val) {
        this->value += (end - begin + 1) * val;
        mark += val;
    }
    
    void maintain() {
        if(begin != end) {
            this->value = left -> value + right->value;
            
        }
        
        
    }
    
    void pushDown(){
        if (mark) {
            
            if(begin!= end) {
                left -> add(mark);
                right -> add(mark);
                
            }
            mark = 0;
        }
    }
    
    int_t query (int begin,int end){
        if(end < this->begin || begin > this->end) return 0;
        if(this->begin >= begin && this->end <= end) return this->value;
        this->pushDown();
        return left->query(begin,end) + right->query(begin, end) ;    
     
     }
     
     void add(int begin,int end,int_t val){
         if(end < this->begin || begin > this->end) return;
         if(this->begin >= begin && this->end <= end) {
             this->add(val);
             return ;
         }
         this->pushDown();
         left->add(begin,end,val);
         right->add(begin,end,val);
         this->maintain();
     }
    
    
};
char buf[400010*sizeof(Node)];
int used = 0;

void* allocate() { return (++used) * sizeof(Node) + buf;}

const int MaxN = 200010;
Node* tree;
int_t a[MaxN];
int n,m;

Node* build(int begin,int end){
    int mid = (begin + end)/2;
    
    Node* node = new(allocate()) Node(begin,end);
    if(begin != end){
        node->left = build(begin ,mid);
        node->right = build(mid+1,end);
        
        
    }
    
    else if(begin == end){
        node->value = a[begin];
    }
    
    node->maintain();
    return node;
}




int main(){
    scanf("%d%d",&n,&m);
    
    for(int i = 1; i <= n; i++){
        scanf("%lld",&a[i]);
    }
    
    tree = build(1,n);
    
    
    for(int i = 1; i <= m; i++){
        int opt,x,y;
        int_t k;
        scanf("%d",&opt);
        if(opt == 1){
            scanf("%d%d%lld",&x,&y,&k);
            tree->add(x,y,k);
            
        }
        else
        {
            scanf("%d%d",&x,&y);
            
            printf("%lld\n",tree->query(x,y));
        }
    }
    
    return 0;
}

Code