UVa

UVa12538 Version Controlled IDE

題目連結

  • 題意：給定 $N$ 筆操作，每筆操作要記錄一個版本，操作有以下幾種：(1)在位置 $p$ 插入 $s$ (2)從位置 $p$ 刪除 $c$ 個字元 (3)輸出版本 $v$ 位置 $p$ 開始的 $c$ 個字元。
  • 題解：持久化 Treap。 
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    #pragma GCC optimize("O2")
    #include <bits/stdc++.h>
    using namespace std;
    using LL = long long;
    using ULL = unsigned long long;
    using PII = pair<int, int>;
    using PLL = pair<LL, LL>;
    using VI = vector<int>;
    using VVI = vector<vector<int>>;
    const int INF = 1e9;
    const int MXN = 3e7;
    const int MXV = 0;
    const double EPS = 1e-9;
    const int MOD = 1e9 + 7;
    #define MP make_pair
    #define PB push_back
    #define F first
    #define S second
    #define FOR(i, L, R) for (int i = L; i < (int)R; ++i)
    #define FORD(i, L, R) for (int i = L; i > (int)R; --i)
    #define IOS                                                                    \
        cin.tie(nullptr);                                                          \
        cout.tie(nullptr);                                                         \
        ios_base::sync_with_stdio(false);
    struct Node
    {
        static Node mem[MXN];
        Node *Lc, *Rc;
        char ch;
        int sz;
        Node() {}
        Node(char _ch) : Lc(nullptr), Rc(nullptr), ch(_ch), sz(1) {}
    } Node::mem[MXN], *ptr = Node::mem;

    inline int sz(Node *a) { return a ? a->sz : 0; }

    inline void pull(Node *a)
    {
        if (!a)
        {
            return;
        }
        a->sz = 1 + sz(a->Lc) + sz(a->Rc);
    }

    void split(Node *t, Node *&a, Node *&b, int k)
    {
        if (t == NULL)
        {
            a = b = NULL;
            return;
        }
        if (sz(t->Lc) + 1 <= k)
        {
            a = new (ptr++) Node(*t);
            split(t->Rc, a->Rc, b, k - 1 - sz(t->Lc));
            pull(a);
        }
        else
        {
            b = new (ptr++) Node(*t);
            split(t->Lc, a, b->Lc, k);
            pull(b);
        }
    }

    Node *merge(Node *a, Node *b)
    {
        if (!a || !b)
        {
            return (a ? a : b);
        }
        Node *res;
        if (rand() % (sz(a) + sz(b)) < sz(a))
        {
            res = new (ptr++) Node(*a);
            res->Rc = merge(a->Rc, b);
        }
        else
        {
            res = new (ptr++) Node(*b);
            res->Lc = merge(a, b->Lc);
        }
        pull(res);
        return res;
    }

    int n;
    Node *ver[MXN];
    string s;
    int d;

    void print(Node *t)
    {
        if (!t)
        {
            return;
        }
        print(t->Lc);
        cout << t->ch;
        if (t->ch == 'c')
        {
            ++d;
        }
        print(t->Rc);
    }

    int main()
    {
        // IOS;
        cin >> n;
        int vi = 0;
        FOR(i, 0, n)
        {
            Node *Lt, *Mt, *Rt;
            int x;
            cin >> x;
            if (x == 1)
            {
                ++vi;
                ver[vi] = ver[vi - 1];
                int p;
                cin >> p >> s;
                p -= d;
                split(ver[vi], Lt, Rt, p);
                FOR(i, 0, s.size()) { Lt = merge(Lt, new (ptr++) Node(s[i])); }
                ver[vi] = merge(Lt, Rt);
            }
            else if (x == 2)
            {
                ++vi;
                ver[vi] = ver[vi - 1];
                int p, c;
                cin >> p >> c;
                p -= d;
                c -= d;
                split(ver[vi], Lt, Rt, p - 1);
                split(Rt, Mt, Rt, c);
                ver[vi] = merge(Lt, Rt);
            }
            else
            {
                int v, p, c;
                cin >> v >> p >> c;
                v -= d;
                p -= d;
                c -= d;
                split(ver[v], Lt, Rt, p - 1);
                split(Rt, Mt, Rt, c);
                print(Mt);
                cout << '\n';
            }
        }
    }

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