Luogu3919 【模板】可持久化数组(主席树)
阅读原文时间:2023年07月08日阅读:2

主席树模板题,注意空间\((n+m) \log(n)\)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

//#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 20000007;
struct Chairman{
    // space complexity : (n + m) * log(n)
    int rt[1000007], T[N], L[N], R[N];
    int treeIndex;
    inline int Build(int l, int r){
        int root = ++treeIndex;
        if(l == r){
            io >> T[root];
            return root;
        }
        int mid = (l + r) >> 1;
        L[root] = Build(l, mid);
        R[root] = Build(mid + 1, r);
        return root;
    }
    inline int Updata(int rt, int l, int r, int x, int w){
        int root = ++treeIndex;
        if(l == r){
            T[root] = w;
            return root;
        }
        L[root] = L[rt], R[root] = R[rt];
        int mid = (l + r) >> 1;
        if(x <= mid) L[root] = Updata(L[rt], l, mid, x ,w);
        else R[root] = Updata(R[rt], mid + 1, r, x, w);
        return root;
    }
    inline int Query(int rt, int l, int r, int x){
        if(l == r) return T[rt];
        int mid = (l + r) >> 1;
        if(x <= mid) return Query(L[rt], l, mid, x);
        else return Query(R[rt], mid + 1, r, x);
    }
}t;
int main(){
    t.treeIndex=0;
    int n,m;
    io >> n >> m;

    t.Build(1,n);
    t.rt[0]=1;

    R(i,1,m){
        int edition,opt;
        io >> edition >> opt;
        if(opt==1){
            int pos, newValue;
            io >> pos >> newValue;
            t.rt[i] = t.Updata(t.rt[edition], 1, n, pos, newValue);
        }
        if(opt==2){
            int pos;
            io >> pos;
            printf("%d\n", t.Query(t.rt[edition], 1, n, pos));
            t.rt[i] = t.rt[edition];
        }
    }

    return 0;
}