• [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
    " , query(root[i] , root[vi] , 1 , n , loc));    
                    } else
                    {
                            int value;
                            read(value);
                            modify(root[i] , root[vi] , 1 , n , loc , value);
                    }            
            }
            
            return 0;
        
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9971287.html
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