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  • 洛谷

    https://www.luogu.org/problem/P4008

    无旋Treap也可以维护序列。
    千万要注意要先判断p节点存在才进行Show操作,不然输出一个''(或者RecBin里面的东西)草。

    假如有限制同时存在的节点数量的话,UnBuild操作是显得重要的。

    当然这里最主要的是类似笛卡尔树的O(n)建立Treap树。

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    #define ls(p) ch[p][0]
    #define rs(p) ch[p][1]
    
    const int MAXN = 2000000 + 5;
    char val[MAXN];
    int ch[MAXN][2], rnd[MAXN], siz[MAXN], tot, root;
    
    int cur;
    stack<int> RecBin;
    
    inline void Init() {
        root = 0, tot = 0;
        cur = 0;
        srand(19260817);
        while(RecBin.size())
            RecBin.pop();
    }
    
    inline void PushUp(int p) {
        siz[p] = siz[ls(p)] + siz[rs(p)] + 1;
    }
    
    void SplitRank(int p, int rk, int &x, int &y) {
        if(!p) {
            x = y = 0;
            return;
        }
        //PushDown(p);
        if(rk <= siz[ls(p)]) {
            y = p;
            SplitRank(ls(p), rk, x, ls(p));
            PushUp(y);
        } else {
            x = p;
            SplitRank(rs(p), rk - siz[ls(p)] - 1, rs(p), y);
            PushUp(x);
        }
    }
    
    int Merge(int x, int y) {
        if(!x || !y)
            return x | y;
        //这个是小根Treap
        if(rnd[x] < rnd[y]) {
            //PushDown(x);
            rs(x) = Merge(rs(x), y);
            PushUp(x);
            return x;
        } else {
            //PushDown(y);
            ls(y) = Merge(x, ls(y));
            PushUp(y);
            return y;
        }
    }
    
    inline int NewNode(char v) {
        int p;
        if(RecBin.size()) {
            p = RecBin.top();
            RecBin.pop();
        } else
            p = ++tot;
        ch[p][0] = ch[p][1] = 0;
        val[p] = v;
        rnd[p] = rand();
        siz[p] = 1;
        return p;
    }
    
    void Show(int p) {
        if(!p)
            return;
        Show(ls(p));
        putchar(val[p]);
        Show(rs(p));
    }
    
    //O(n)建树,返回新树的根
    int st[MAXN], stop;
    char buf[MAXN];
    inline int Build(int n) {
        stop = 0;
        for(int i = 0; i < n; ++i) {
            int tmp = NewNode(buf[i]), last = 0;
            while(stop && rnd[st[stop]] > rnd[tmp]) {
                last = st[stop];
                PushUp(last);
                st[stop--] = 0;
            }
            if(stop)
                rs(st[stop]) = tmp;
            ls(tmp) = last;
            st[++stop] = tmp;
        }
        while(stop)
            PushUp(st[stop--]);
        return st[1];
    }
    
    //O(n)回收整棵树
    inline void UnBuild(int p) {
        if(!p)
            return;
        UnBuild(ls(p));
        UnBuild(rs(p));
        RecBin.push(p);
    }
    
    inline void Move() {
        scanf("%d", &cur);
    }
    
    inline void Insert(int &root) {
        int x = 0, y = 0, z = 0, n;
        SplitRank(root, cur, x, z);
        scanf("%d", &n);
        buf[n] = '';
        for(int i = 0; i < n; ++i) {
            char ch = getchar();
            while(ch < 32 || ch > 126 || ch == '
    ' || ch == '
    ')
                ch = getchar();
            buf[i] = ch;
        }
        y = Build(n);
        root = Merge(Merge(x, y), z);
    }
    
    inline void Delete(int &root) {
        int x = 0, y = 0, z = 0, n;
        SplitRank(root, cur, x, y);
        scanf("%d", &n);
        SplitRank(y, n, y, z);
        //会不会太慢了
        UnBuild(y);
        //y=Merge(ls(y),rs(y));
        root = Merge(x, z);
    }
    
    inline void Get(int &root) {
        int x = 0, y = 0, z = 0, n = 0;
        SplitRank(root, cur, x, y);
        scanf("%d", &n);
        SplitRank(y, n, y, z);
        Show(y);
        puts("");
        root = Merge(Merge(x, y), z);
    }
    
    inline void Prev() {
        --cur;
    }
    
    inline void Next() {
        ++cur;
    }
    
    char op[20];
    int main() {
    #ifdef Yinku
        freopen("Yinku.in", "r", stdin);
    #endif // Yinku
        int t;
        scanf("%d", &t);
        Init();
        while(t--) {
            scanf("%s ", op);
            switch(op[0]) {
                case 'M':
                    Move();
                    break;
                case 'I':
                    Insert(root);
                    break;
                case 'D':
                    Delete(root);
                    break;
                case 'G':
                    Get(root);
                    break;
                case 'P':
                    Prev();
                    break;
                case 'N':
                    Next();
                    break;
            }
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Yinku/p/11330315.html
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