[模板] 可持久化数组

[题目链接]

         https://www.luogu.org/problemnew/show/P3919

[算法]

       可持久化线段树模板

       详见这篇文章 : https://www.cnblogs.com/RabbitHu/p/segtree.html

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e6 + 10;

int n , m , version , idx;
int a[MAXN] , sum[MAXN << 4] , root[MAXN << 4] , lson[MAXN << 4] , rson[MAXN << 4];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void build(int &k , int l , int r)
{
        k = ++idx;
        if (l == r) 
        {
                sum[k] = a[l];
                return;        
        }
        int mid = (l + r) >> 1;
        build(lson[k] , l , mid);
        build(rson[k] , mid + 1 , r);
        sum[k] = sum[lson[k]] + sum[rson[k]];
}
inline void modify(int &k , int old , int l , int r , int pos , int value)
{
        k = ++idx;
        lson[k] = lson[old] , rson[k] = rson[old];
        if (l == r) 
        {
                sum[k] = value;
                return;
        }
        int mid = (l + r) >> 1;
        if (mid >= pos) modify(lson[k] , lson[k] , l , mid , pos , value);
        else modify(rson[k] , rson[k] , mid + 1 , r , pos , value);    
        sum[k] = sum[lson[k]] + sum[rson[k]];  
}
inline int query(int &k , int old , int l , int r , int pos)
{
        k = ++idx;
        lson[k] = lson[old] , rson[k] = rson[old];
        sum[k] = sum[old];
        if (l == r) return sum[k];
        int mid = (l + r) >> 1;
        if (mid >= pos) return query(lson[k] , lson[k] , l , mid , pos);
        else return query(rson[k] , rson[k] , mid + 1 , r , pos);
}

int main()
{
        
        read(n); read(m);
        for (int i = 1; i <= n; i++) read(a[i]);
        build(root[version = 0] , 1 , n);
        for (int i = 1; i <= m; i++)
        {
                int vi , type , loc;
                read(vi); read(type); read(loc);
                if (type == 2)
                {
                        printf("%d\n" , query(root[i] , root[vi] , 1 , n , loc));    
                } else
                {
                        int value;
                        read(value);
                        modify(root[i] , root[vi] , 1 , n , loc , value);
                }            
        }
        
        return 0;
    
}